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 $$H : I \to \text{Hom}(C, D)$$
such that the restriction of the map to $0$ is $f$ and the restriction of the map to $1$ is $g$, where $\text{Hom}$ is the hom chain complex. 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 $I$.
Note that in the topological case it's pretty clear that topological spaces, continuous functions, and homotopies don't form a 2-category for the same reason that closed loops in a space based at a point don't form a group (composition is only associative up to homotopy).