Proof Explanation in Hatcher, Theorem 3.43 Here is the link to the book: http://pi.math.cornell.edu/~hatcher/AT/AT.pdf
In Theorem 3.43, checking the commutativity of the diagram is left to the reader. The first square commutes trivially, but I have no idea of the second and the third  squares. I also can't see where the $[B]$ in the diagram came from. Thanks in advance. 
Edit. This theorem is a generalization of a theorem in Bredon's Topology and Geometry, Theorem VI.9.2, and the proof is similar too. I understood the proof in Bredon's but I think it is nontrivial to apply the same proof directly.
 A: OK I just tried to work through this so here's my attempt at commutativity. Actually as Bredon points out, the square only commutes up to sign, but that's actually good enough.
We asume that $A$ and $B$ keep the orientation of $\partial M$.
Pick $[\phi] \in H^k(M, A)$, where $\phi \in C^k(X)$ is zero on $C_k(A)$. I'll assume that the fundamental classes $[M] \in H^k(M, \partial M), [A] \in H^k(A, \partial A)$ and $[A] \in H^k(A, \partial A)$ are represented by cycles $\mu \in C^k(M), \beta \in C^k(A)$ and $\alpha \in C^k(A)$ respectively.
We observe that $\alpha + \beta$ represents $[\partial M]$ : First note that $[\partial \left( \alpha + \beta\right)] = [\partial A] - [\partial A]$ (the orientations of $\partial A$ and $\partial B$ differ by a sign) so $[A] + [B]$ indeed represent a class in $H_{n-1}(\partial M)$. Second, this class restricts to a local orientation of $\partial M$ on any interior point of $A$ or $B$, so by the isomorphism $H_{n-1}(\partial M) \cong H_{n-1}(\partial M , \partial M - x)$ for $x \in \partial M$ this class represents $[\partial M]$.
We denote the lower horizontal map in the relevant square $\delta$. Now lets take a look at the result of going down first and then right
$$
\delta ([M] \frown [\phi]) = [ \partial (\mu \frown \phi ) ] = [ (-1)^k (\partial \mu \frown \phi - \mu \frown \delta \phi) ] = (-1)^k  [ \partial M ] \frown [\phi]
$$
(since $\delta \phi = 0$). Now we consider going right first and then down (in two steps). The first two maps only consist of restricting $\phi$, first to $C_k(\partial M)$ and the to $C_k(B)$, so the result of this map is
$$
[B] \frown \left[\phi|_{C_k(B)}\right] = [\beta \frown \phi] = [(\alpha + \beta) \frown \phi] = [\partial M] \frown [\phi]
$$
And we are done.
