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 interval object 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).
With this definition one can work guided by analogy to 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. I think everything reduces to working with fairly concrete maps between some chain complexes constructed from $I$.