Checking the homomorphism $f: \mathbb{R}/\mathbb{Z} \to S^1$ is well-defined?

The mapping $[0,1)\to S^1$, $t\mapsto e^{2\pi i t}$ is well-defined, and assuming you've already seen some basic complex analysis there's no need to go deeper into this. But for the sake of completeness, here goes. One typically defines the exponential as the mapping $\Bbb{C}\to\Bbb{C}$, $z\mapsto\sum_{n=0}^{\infty}\frac{z^n}{n!}$. One can obviously restrict the domain to get a mapping $[0,1)\to\Bbb{C}$, $t\mapsto e^{2\pi i t}$. The only question is whether $e^{2\pi i t}$ actually lies in $S^1$, and of course the answer is yes (clear once you know $e^{a+b}=e^ae^b$ and $\overline{e^z}=e^{\overline{z}}$, since then $|e^{2\pi i t}|^2=e^{2\pi i t}e^{-2\pi i t}=e^0=1$).

Now, if you define $\Bbb{R}/ \Bbb{Z}$ to be the set $[0,1)$, then there's nothing more involved in showing $f$ is a well-defined function. Having said this, one typically defines $\Bbb{R}/ \Bbb{Z}$ as a quotient group, so set theoretically it's not equal to $[0,1)$. Of course it's very trivial to establish the bijection $\Bbb{R}/\Bbb{Z}\cong [0,1)$; so if you start with this quotient definition of $\Bbb{R}/\Bbb{Z}$, then showing that $f$ is a well-defined mapping takes just slightly more effort (actually this is a matter of complex analysis, in that it uses the periodicity of the 'imaginary' exponential).

A separate question which arises (though you didn't ask) is whether or not $f$ is actually a homomorphism from the abelian group $\Bbb{R}/\Bbb{Z}$ into $S^1$ (does it preserve the group operations). Proving this directly is not too bad though you may have to deal with some small amount of case work (regardless of which definition you adopt). But the standard way of proving that a map from a quotient group to some other group is a homomorphism is to first consider a map from the not-quotiented group, and then investigate its kernel. Explicitly, consider $F:\Bbb{R}\to S^1$, $F(t)=e^{2\pi i t}$. This is a well-defined mapping, and it's a homomorphism because of the properties of complex exponentials. The $2\pi$-periodicity of the exponential (a complex-analysis fact) means that $\ker F=\Bbb{Z}$. Therefore, (universal property of quotient groups) there is a well-defined group homomorphism $f:\Bbb{R}/\Bbb{Z}\to S^1$ such that for all $t\in \Bbb{R}$, we have $F(t)= f([t])$, where $[t]\in\Bbb{R}/\Bbb{Z}$ denotes the equivalence class.

Similar remarks in the $\Bbb{Z}/n\Bbb{Z}$ case. If set theoretically you define this to be $\{0,\dots, n-1\}$, there's nothing to be checked in terms of well-definition of functions having this as the domain, because there are no equivalence classes to be dealt with in this approach. What does require a small verification is the equivalence between this definition and the usual equivalence-class definition of $\Bbb{Z}/n\Bbb{Z}$. Also, typically, we're not just interested in the functions with these domains; we're interested in whether or not they're group homomorphisms (and for that the standard way is the one I described above, using the universal property).