Part of the problem is that you're not using a convenient definition of chain homotopy. I'll use a differential going down in degree. Let $I$ be the <a href="http://ncatlab.org/nlab/show/interval+object">interval object</a> in the category of chain complexes; that is, $I$ is $\text{span}(0, 1)$ in degree $0$ and $\text{span}(e)$ in degree $1$ ($k$ the underlying commutative ring), and the differential sends $e$ to $1 - 0$. If $C, D$ are two complexes, I claim that a chain homotopy between two chain maps $f, g : C \to D$ is precisely a chain map
$$H : C \otimes I \to D$$

such that the restriction of the map to $C \otimes 0$ is $f$ and the restriction of the map to $C \otimes 1$ is $g$ (it may be necessary to put $I$ on the other side depending on your conventions for tensor products). This is exactly analogous to the topological situation.

With this definition one can work exactly as in the topological situation (there the 2-category is the 2-category of topological spaces, continuous functions, and homotopy classes of homotopies between such functions); all of the maps you need have obvious topological definitions, although I admit I have never worked out the details.