Center and Commutator of Profinite Groups

You can calculate the commutator of an element from the centraliser of its reductions in a very general case. Let $R$ be a ring that is adically separated with respect to an ideal $I$ (so $\bigcap_{i=k}^\infty I^k = 0$). Given two matrices $A, B \in \mathrm{M}_n(R)$, we denote their reductions to $\mathrm{M}_n(R / I^k)$ by $A_k$ and $B_k$. Then $AB = BA$ if and only if $A_kB_k = B_kA_k$ for all $k$. The implication $\Rightarrow$ is clear, since $R \to R / I^k$ is a ring homomorphism. For $\Leftarrow$, note that $AB - BA$ reduces to $A_kB_k - B_kA_k = 0 \in \mathrm{M}_n(R)$ for all $k$. This means $AB - BA \in \mathrm{M}_n(I^k)$ for all $k$. Since $R$ is $I$-adically complete, this means $AB = BA$.

From this we can deduce that for every subgroup $G \leq \operatorname{Gl}_n(R)$ and every element $x \in G$, the centraliser $\mathrm{Z}_G(x)$ is the intersection of all preimages $\pi_k^{-1}(\mathrm{Z}_{G_k}(x_k))$. Here, $pi_k : \operatorname{Gl}_n(R) \to \operatorname{Gl}_n(R / I^k)$ is the reduction map, $G_k = \pi_k(G)$, and $x_k = \pi_k(x)$. To check this statement, let $y \in G$. Then $y \in \mathrm{Z}_G(x)$ means $xy = yx$, which is equivalent to $x_ky_k = y_kx_k$ for all $k$, i.e. $y_k \in \mathrm{Z}_{G_k}(x_k)$ for all $k$. In other words, $y \in \pi_k^{-1}(\mathrm{Z}_{G_k}(x_k))$.

Now for your concrete case: We can pick $R = \mathbb{F}_q[[t]]$, $I = (t)$, and $G = \mathrm{UI}_n$. Then $G_k = \mathrm{UI}_n^{(k)}$. Then for $x \in \mathrm{UI}_n$, we get $$ \mathrm{Z}_{\mathrm{UI}_n}(x) = \bigcap_{k = 1}^\infty\pi_k^{-1}(\mathrm{Z}_{\mathrm{UI}_n^{(k)}}(x_k)) = \varprojlim_k \mathrm{Z}_{\mathrm{UI}_n^{(k)}}(x_k) \leq \mathrm{UI_n}. $$