This question was essentially answered in the comments. I'm collecting the information in an answer, because it might be interesting to some people.

Firstly, the lemma as stated is false. There was some discussion about whether the lemma can be iterated. To see this is the case, note a simplicial collapse of B onto a subcomplex containing $A$ in particular must fix $A$ and will be homotopy equivalence rel $A$. Using iteration the following simplicial complexes $B$ with $A = \ast$ give counterexamples:

- A triangulation of the house with two rooms is a contractible but not collapsible simplicial complex of dimension 2.
- The second barycentric subdivision of the dunce hat is another contractible but not collapsible simplicial complex of dimension 2.
- Rudin's triangulation is a non-collapsible triangulation of the 3-disk.
- This question lists various other interesting examples of non-collapsible triangulations of disks.

Without iterating you get a counterexample from Higman's group (or any other finitely-presented acyclic group). This is the finitely presented acyclic group $G = \langle x_i|x_ix_{i+1}x_i^{-1}\rangle_{i \in \mathbb{Z}/4\mathbb{Z}}$. There is a finite simplicial complex $BG$ of dimension 2 that is a classifying space for $G$. Taking $A = *$, we get that $H^2(BG,*;\mathbb{Z}/2\mathbb{Z}) = 0$, so the lemma would imply that $BG$ collapses onto, and thus is homotopy equivalent to, a simplicial complex of dimension 1. But every finite simplicial complex of dimension 1 has a free group as fundamental group.

As Oscar Randal-Williams suggests in the comments, the lemma is true if one additionally assumes that at most two $m$-simplices in $B\backslash A$ attach to an $(m-1)$-simplex in $B \backslash A$. For suppose this is the case, but all $m$-simplices $S$ in $B \backslash A$ that have the property that each face not in $A$ is the face of another $m$-simplex (there is then exactly one of these by the additional assumption). It suffices to prove we can collapse at least one $m$-simplex in $B \backslash A$, because the additional assumption and homological condition are preserved by such a collapse. Consider the sum of all $m$-simplices in the cellular chain complex of the pair $(B,A)$: $\sigma = \sum_i S_i$. Then $\partial \sigma$ is $0$, as faces either lie in $A$ or appear twice (and we are working modulo $2$). Thus $\sigma$ is a cycle, but this contradicts $H_m(B,A;\mathbb Z/2\mathbb Z)=0$, which follows from the assumption on cohomology using the universal coefficient theorem.

This additional assumption is satisfied if $B$ is a PL manifold of dimension $m$, because in that case the link of every $(m-1)$-simplex is either $S^0$ or a point (depending on wither the $(m-1)$-simplex is in the boundary or not). One cannot iterate the lemma with the additional assumption, so we do not get contradiction (e.g. from Rudin's triangulation). Coincidentally, it does prove the classical result that every triangulation of the 2-disk is collapsible (apply the lemma with additional assumption, and then note that every tree is collapsible).

Finally, if one looks to Lemma (e) on page 257 of Kirby-Siebenmann and its eventual application in the proof of Theorem A.1 on page 262, part ($\delta$), we see that in the proof we can get away with only using the lemma in the case that $B$ is a PL-manifold of dimension $m$. The sliced structure classification theorem for open 4-manifolds that do not admit a handle decomposition is safe.

union a bunch of stuff $B'$ in the $(m-1)$ skeleton. In order to "iterate this lemma", you would require the reduced homology of that $B'$ to also be zero, no? $\endgroup$ – Vidit Nanda Oct 22 '14 at 18:52