Why are cochains in group cohomology exact as a functor of the coefficients?

As Alex Youcis commented, you can understand this much more cleanly using inhomogeneous cochains. Given a continuous map $F:G^n\to A$, you can define $f:G^{n+1}\to A$ by $$f(g_0,g_1,\dots,g_n)=g_0F(1,g_0^{-1}g_1,\dots,g_0^{-1}g_n)$$ and you can easily check that $f\in C^n(G,A)$. Conversely, given $f\in C^n(G,A)$, you can define $$F(g_1,\dots,g_n)=f(1,g_1,\dots,g_n)$$ and this is inverse to the previous construction. This gives a natural isomorphism of abelian groups between $C^n(G,A)$ and the set of all continuous maps $F:G^n\to A$. But this latter construction is trivially seen to be exact (for right-exactness, note that if $p:B\to C$ is a surjection of $G$-modules and $F:G^n\to C$ is continuous, we can compose $F$ with any set-theoretic section of $p$ to get a continuous map $G^n\to B$).

Another way to think about what's going on here is that the topological $G$-set $G^{n+1}$ is actually freely generated by the subset $\{1\}\times G^n$. So, $G$-equivariant continuous maps out of $G^{n+1}$ are the same thing as just any continuous maps out of $\{1\}\times G^n$.