Let me complement Thiago's answer with what we said in the comments below. This is just a very long comment.

Note that if we restrict our attention to extremally disconnected sets, the situation is much nicer. Indeed, $Cond(Ab) \simeq Fun^\times(ExDisc^{op}, Ab)$, where $Fun^\times$ is the category of finite product preserving functors.

In particular, because in $Ab$, finite products are coproducts, the inclusion $Fun^\times(ExDisc^{op},Ab) \subset Fun(ExDisc^{op},Ab)$ preserves all colimits, and because in the latter, colimits are pointwise, the same holds for the later. In particular, if $S$ is extremally disconnected, the formula $(colim_i F_i)(S) = colim_i F_i(S)$ always holds for diagrams of condensed abelian groups (in particular, for cokernels).

In general, you have to sheafify, including for cokernels.

However, there are special cases where you can forget about it : for instance, as proved in theorem 3.2 of Scholze's notes, if $A$ is discrete, then $H^1(S,A) = 0$ for $S$ profinite too, so the formula $coker(A\to B)(S) = coker(A(S)\to B(S))$ holds also for profinite $S$.

This formula also holds if $S$ is profinite and $A,B$ are *solid* (which is a kind of "completion" requirement). Indeed, solid modules are closed under colimits, so the cokernel will be solid too, and $\mathbb Z[S]^\blacksquare$ is projective in solid modules. This implies the claim as by definition, solidity implies that $\hom(\mathbb Z[S]^\blacksquare, C) \cong C(S)$ .

In fact, as Zhouhang points out in the comments below this answer, lemma 5.9 in the condensed notes shows that solid implies "derived solid", and so in particular $H^1(S,A) = 0$ if $A$ is solid and $S$ profinite, so really it suffices to assume $A$ is solid. I prefer to leave the previous argument as is, because it's easier to prove.

Now let $X$ be a profinite set. Projectivity of $\mathbb Z[X]$ amounts exactly to $H^1(X,-)$ vanishing, and so to the coker formula holding in full generality when evaluating at $X$. Peter Scholze mentioned in the comments that there *are* nontrivial examples of such a situation, i.e. where $\mathbb Z[X]$ is projective but $X$ is not extremally disconnected. These apparently always have a dense open subset which is extremally disconnected, and so that rules out many (most ?) examples.

The situation is even worse if you try to evaluate at a general compact Hausdorff space, and you should definitely not expect it to commute with colimits, or even specifically cokernels.