Well you're looking for a map $A^{n+1}\oplus B^n\to A^{n+1}\oplus B^n$, and your data is a homotopy $f\to g$, that is, a whole bunch of maps $h_n : A^{n+1}\to B^n$.
So you might try to use that.
How about $K:(a,b) \mapsto (a,b+h(a))$ (forgetting the indices) ?
Then let's see how it interacts with the differentials : $d_f(K(a,b)) = d_f(a,b+h(a)) = (d(a), d(b+h(a))+f(a)) = (d(a), d(b)+dh(a) + f(a))$
Now say you have something like $dh+hd = g-f$, so that $dh+f = g-hd$, you get $(d(a), d(b)+g(a) - hd(a))$
Compare with $K(d_g(a,b)) = K((d(a), d(b)+g(a))) = (d(a), d(b)+g(a)+hd(a))$
That's almost the same thing, up to a sign.
Well this is not a problem : it's just that I most likely messed up some of the sign conventions (for instance, $d$ on $A[1]$ is probably something like $-d_A$ if you pay more attention).
So this is just a rough sketch of what it'll look like, but what I'm saying is that $K$ (or some slight modification of it) will probably be a chain map between $C(g)$ and $C(f)$
Now there will be a similar chain map from $C(f)$ to $C(g)$, and with a bit of luck it'll be easy to show that the two are homotopy inverses to one another.
One approach is to use the mapping cone of a chain complex. For some details, see these notes, for example, and in particular Corollary 2.5 is getting to the result you want. Addendum: the cited notes are really just replicating section 1.5 of Weibel's book An Introduction to Homological Algebra. Corollary 1.5.4 is the relevant one.
Given a chain map $f: C_{\bullet} \to D_{\bullet}$, define its mapping cone $C(f)_{\bullet}$ to be the chain complex with $n$th term $C_{n-1} \oplus D_{n}$, boundary map $\begin{pmatrix} -\partial^{C} & 0 \\ f & \partial^{D} \end{pmatrix}$. Then there is a natural map $D_{\bullet} \to C(f)_{\bullet}$ and the maps $C_{\bullet} \to D_{\bullet} \to C(f)_{\bullet}$ induce a long exact sequence in homology. If the map $f$ is a quasi-isomorphism, then $C(f)_{\bullet}$ is free (that's the corollary from the notes), and furthermore it is degree-wise free. If it is bounded below, then it must be contractible (chain-homotopy equivalent to zero). This is not hard to prove by induction. (See also https://mathoverflow.net/questions/73687/when-mapping-cone-is-contractible and https://mathoverflow.net/questions/73687/when-mapping-cone-is-contractible, which both at least peripherally address the case of chain complexes, in addition to topological spaces.)
Now I claim that the chain homotopy between $1$ and $0$ on $C(f)_\bullet$ gives a chain homotopy equivalence between $C_\bullet$ and $D_\bullet$. This requires some checking.
Best Answer
Yes there is; I will assume you're familiar with some homology of spaces, say singular homology. Recall how we construct singular homology for a space $X$ is, we take the abelian groups $C_n(X)$ to be the free abelian group generated the set $\{\text{continuous maps }\Delta^n\to X\}$, and these form a chain complex $$\cdots\to C_{n+1}(X)\to C_n(X)\to C_{n-1}(X)\to\cdots$$ Furthermore, if $f:X\to Y$ is a continuous map, then we get maps $f_n:C_n(X)\to C_n(Y)$ by taking a map $\sigma:\Delta^n\to X$ to the composition $f\circ\sigma:\Delta^n\to Y$, and these in fact give a morphism of chain complexes $f_*:C_*(X)\to C_*(Y)$.
Now, suppose we have two maps $f,g:X\to Y$ inducing morphisms of chain complexes $f_*,g_*:C_*(X)\to C_*(Y)$, and suppose we have a homotopy $h:X\times I\to Y$ from $f$ to $g$. Then we can use this to construct $h_n:C_n(X)\to C_{n+1}(Y)$; this is not trivial and I won't put the entire thing here, but basically given some map $\sigma:\Delta^n\to X$, we have the composition $h\circ(\sigma\times\mathrm{id}_I):\Delta^n\times I\to Y$, and you can subdivide $\Delta^n\times I$ into (n+1)-simplices in a way that allows you to identify this map $\Delta^n\times I\to Y$ with a formal sum of maps $\Delta^{n+1}\to Y$ (you can find the details in Hatcher). All together, these $h_n:C_n(X)\to C_{n+1}(Y)$ can be shown to give a homotopy between $f_*$ and $g_*$ in the sense of chain complexes, giving the connection you seek.