I am not really sure about the scope of this question and the type of answer OP is looking for but computing those higher spin matrices/representations successively is relatively straight forward:

Lets assume for now that we have understood the spin $\frac{1}{2}$ case: so we know that the spin $\frac{1}{2}$ operators are $$ \begin{align} S_x = \frac{\hbar}{2} \sigma_x= \frac{\hbar}{2}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, \quad S_y = \frac{\hbar}{2} \sigma_y= \frac{\hbar}{2}\begin{pmatrix} 0 & -\mathrm{i}\\ \mathrm{i} & 0 \end{pmatrix}, \quad S_z = \frac{\hbar}{2} \sigma_z= \frac{\hbar}{2}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}. \end{align} $$ We can verify that those sattisfy the defining commutator relation $\left[ S_j, S_k\right] = i \hbar \varepsilon_{jkl} S_l$ with $i,j,k\in\{x,y,z\}$. Furthermore we can compute the Casimir operator $S^2$: $$ S^2=S_x^2+S_y^2+S_z^2 =\hbar^2\begin{pmatrix} \frac{3}{4} & 0\\ 0 & \frac{3}{4} \end{pmatrix}. $$ The eigenvalues of $S^2$ characterize the present representation and are $\hbar^2\frac{3}{4}=\hbar^2s(s+1)$ with $s=\frac{1}{2}$. A spin state can be completely characterized by specifying $s$ and one additional spin projection which is conventionally chosen to be along the $z$-direction. $S_z$ has two Eigenvalues $m_\frac{1}{2}=\pm\frac{\hbar}{2}$. The simultaneous Eigenvectors of $S^2$ and $S_z$: $(1,0)^T$ and $(0,1)^T$ form the basis of spin states with well defined quantum numbers $s$ and $m_s$. Eigenvectors of the other spin operators $S_x$ and $S_y$ can be computed and expressed in this basis.

The quoted wikipedia article states that by taking Kronecker products of the spin $\frac{1}{2}$ representation with itself repeatedly, one may construct all higher irreducible representations. What this means in practice is that we can use the Kronecker products to couple spins. The important point here is that when we couple spins the naive basis for states which we get from the direct product of spin states is not an Eigenbasis of the coupled spin operators. Let me illustrate this with an example. We begin with coupling two spin $\frac{1}{2}$: $$ \tilde{S_i}=S_i\otimes \mathrm{Id}_2+\mathrm{Id}_2\otimes S_i $$ which results in $$ \tilde{S}_x=\frac{\hbar}{2} \begin{pmatrix}0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{pmatrix},\quad \tilde{S}_y=\frac{\hbar}{2\mathrm{i}} \begin{pmatrix} 0 & 1 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ 0 & -1 & -1 & 0 \end{pmatrix},\quad \tilde{S}_z=\hbar\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}. $$ $\tilde{S}_z$ has four eigenvalues: $+1$, $0$, $0$ and $-1$. To fully characterize the coupled states we need to compute the Casimir operator $\tilde{S}^2$: $$ \tilde{S}^2 = \tilde{S}_x^2+\tilde{S}_y^2+\tilde{S}_z^2=\hbar^2 \begin{pmatrix}2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}. $$ Using the Eigenvectors of $\tilde{S}^2$ we can construct a matrix $U$ which diagonalizes $\tilde{S}^2$: $$ U= \begin{pmatrix} 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 1 & 0 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \Rightarrow U \tilde{S}^2 U^\dagger=\hbar^2 \begin{pmatrix}0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix} $$ We can now classify the coupled Eigenstates of $\tilde{S}^2$ by their eigenvalues: we have three "triplet" states with $s=1$ ($s(s+1)=2$) and one "singlet" state with $s=0$. A representation is classified by the eigenvalue of $\tilde{S}^2$. Using $U$ we can split $\tilde{S}_i$ into block diagonal matrices where each block corresponds to one representation with distinct $s$: $$ U\tilde{S}_xU^\dagger=\hbar \left( \begin{array}{c|ccc} 0 & 0 & 0 & 0 \\ \hline 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\ \end{array} \right),\quad U\tilde{S}_yU^\dagger=\hbar \left( \begin{array}{c|ccc} 0 & 0 & 0 & 0 \\\hline 0 & 0 & -\frac{i}{\sqrt{2}} & 0 \\ 0 & \frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\ 0 & 0 & \frac{i}{\sqrt{2}} & 0 \\ \end{array} \right),\quad U\tilde{S}_zU^\dagger=\hbar\left( \begin{array}{c|ccc} 0 & 0 & 0 & 0 \\\hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ \end{array} \right). $$ The $s=1$ blocks of $U\tilde{S}_i U^\dagger$ are the operators of the $s=1$ representation. Due to properties of the Kronecker product and the block diagonal form of $U\tilde{S}_i U^\dagger$ it is clear that the blocks satisfy the defining commutator relation separately thus we succeeded in constructing a $s=1$ representation by coupling two $s=\frac{1}{2}$ spins/representations and computing the Eigenvectors of the Casimir operator of the coupled spins. Higher representations can be computed by coupling more spins/representations: e.g. coupling $s=1$ with $s=\frac{1}{2}$ will yield the $s=\frac{3}{2}$ quartet (and in this case one additional $s=\frac{1}{2}$ doublet).

Giving a completely precise definition of everything would take a large amount of time, and would likely not be helpful. So instead here I will spell out the general picture, and any terms which are unfamiliar should be independently researched.

Quantum field theory has as its Hilbert space the physical states of the quantum field, see also a Fock Space. Suppose we have a Lagrangian scalar which specifies the equations of motion for the evolution of the quantum field. Demanding Lorentz invariance in a Lagrangian then leads to the natural question of what types of fields can exist that satisfy a "Lorentz transformation law," or that leave the Lagrangian scalar unchanged under an arbitrary Lorentz transformation. In other words, we are interested in (finite) $n-$dimensional representations of the Lorentz group.

The Lorentz group is a continuous Lie group, so we can study the representations by looking at the Lie algebra. It turns out that $L \cong SU(2) \times SU(2)$, and any representation of the Lorentz group can be built from irreducible representations of $SU(2).$ The Casimir Operators of a Lie Algebra are something important to understand here as well.

In ordinary quantum mechanics, the spin is just its own Hilbert space that is direct-producted on to the original Hilbert space of the position/momentum. The spin Hilbert space is an $n-$ dimensional representation of $SU(2),$ which in ordinary quantum mechanics meshes well with rotations because of the correspondence between $SO(3)$ and $SU(2).$ The reason that the stuff towards the beginning of this post isn’t spelled out in introductory texts to QM (like the wikipedia article you linked) is because to actually start doing physical calculations in non relativistic QM you don’t need to know the precise details.

Thanks to all the contributions I think I finally solved the problem. It is similar to problems structuralist mathematicians often have when talking to physics. Stackexchange is full of questions which boil down to similar troubles. I shall first explain the trouble in a banal setting and will move on to my quantum physics problem and what I have learned from it hoping this might be helpful for others.

Banal Setting: Geometry

We are in the real vector space ${\mathbb R}^2$ and look at two tuples of vectors.

${\mathbf a} = \left( \begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix} 0 \\ 1\end{pmatrix} \right)$ and ${\mathbf b} = \left( \begin{pmatrix}1 \\ 7\end{pmatrix}, \begin{pmatrix} 2 \\ 3\end{pmatrix} \right)$.

Do we have an orthonormal basis somewhere?

Physicist: Yes. ${\mathbf a}$ is an orthonormal basis and ${\mathbf b}$ is not.

Structuralist: What do you mean with orthonormal basis? We are in a real vector space. This is an algebraic structure. Both, ${\mathbf a}$ and ${\mathbf b}$ are indeed bases, which is a notion defined in any vector space. However, I do not have the faintest idea what you are talking about when your are asking for an orthonormal basis.

Theorist: Well. Depends. On ${\mathbb R}^2$ it is possible to introduce not only the structure of a real vector space but also the structure of an inner product space. Often, implicitly, without mentioning this, we use the inner product $\begin{pmatrix}x_1 \\ y_1\end{pmatrix} \cdot \begin{pmatrix}x_2 \\ y_2\end{pmatrix} = x_1 \cdot y_1 + x_2 \cdot y_2$. So if we are in this particular inner product space then ${\mathbf a}$ is an orthonormal basis.

Problem: Initially we said: "We are in the real vector space ${\mathbb R}^2$". On this real vector space we can define an inner product in such a way that ${\mathbf b}$ is an orthonormal basis. We can also define an inner product in such a way that none of the two is an orthonormal basis. We must explicitly introduce an additional structural element in order to talk about orthonormal bases. In a real vector space we simply cannot talk about orthonormality, about distances or angles, simply because the structure we need is not there. Of course, we can introduce all kinds of additional structures on ${\mathbb R}^2$, for example we could introduce holomorphic, topologic, differentiable and many more structures, but this amounts to an additional, arbitrary choice which I must make and announce.

My Quantum Physical Problem

My problem was that I believed in the mantra "Quantum physics is done in a Hilbert space".

Since a spin 3/2 measurement can have 4 different values I was looking at Hilbert space ${\mathbb C}^4$.

"Quantum physics is done in a Hilbert space" is correct only for $n$-level systems. When I want to look at spin I need an additional structure identifying spin in this Hilbert space. This is similar to above question of orthonormality, which cannot be dealt with in a vector space but needs an inner product space as additional structure.

When I study spin then I study observables which have the structure of the 3-dimensional Lie algebra $su(2)$. Therefore I am interested in "finding" this structure in the observables of a Hilbert space. Similarly as above, in the case of the inner product, this additional structure can be found in an algebra of observables in different ways, some of which might be equivalent. However, I must introduce this additional structure and define what it means that they are equivalent.

A representation of $su(2)$ in the space of observables or of $SU(2)$ in the unitary group of the vector space provides the structure; equivalency means isomorphism of the representation.

So, spin quantum physics is done in a Hilbert space together with a representation.

When studying these representations one finds, that $SU(2)$ has many representations. They can be classified and their classification then leads to structurally different variants of spin, as in spin 0, spin 1/2, spin 1 etc.

Irreducible representations have no special magic, they are simply "easier" in the following sense: I can do 2-dimensional geometry in a ${\mathbb R}^100$ with standard inner product. The additional 98 dimensions are not really necessary and I can simply cut them away. When I am using a reducible representation I also can cut away additional dimensions of the Hilbert space and restrict myself to the invariant subspace of the representation.

Important are the isomorphism types of the representations.

The Wikipedia article provides matrices for $S_x$ for higher spins. The article makes implicit assumptions on the representation of $SU(2)$. These assumptions are encoded implicitly (1) in the representation of $SU(2)$ in ${\mathbb C}^2$, which is generic, and (2) in the manner of generating new representations in higher dimensions ${\mathbb C}^{> 2}$ using tensor products of representations (or Kronecker products of the associated matrices). They correspond to a choice of a specific representation of the isomorphism type required for the necessary spin type 1, 3/2 or whatever.

So, of course, completely different $S_x$ are possible, but they then also correspond to a different choice of the representation.

Conclusion

My problem with the article was of the same nature as the problem of the structuralist who was calling for an inner product as additional structure.

It is the representation which "glues" the physical meaning to the elements of the Hilbert space.

In this context also "a spinor is an element of a representation" completely makes sense. Of course, every element of the range of a possible representation could play that role, and so the statement looked empty to me (as in: Every vector space basis can play the role of an orthonormal base). However, if we start with a given representation than the choice of an element is a clear structural statement (just as in a specific inner product space the choice of an orthonormal base is not arbitray - provided an inner product has been fixed first).

As pleasant side effect I can now also appreciate the meaning of the Stone-von Neumann theorem and its commutator relations.