Why chain homotopy when there is no topology in the background?

Here's one way to look at it: There is a chain complex $Hom(C,D)$ in which the $n$th chain group is the product over $k$ of $Hom(C_k,D_{n+k})$ and the boundary is given by $\partial (f(c))=(\partial f)(c)+(-1)^{|f|}f(\partial c)$. A chain map is a $0$-cycle, and two of them are chain homotopic if they differ by a boundary.

EDIT: And then a chain map $B\to Hom(C,D)$ corresponds precisely to a chain map $B\otimes C\to D$, where $B\otimes C$ is defined using the usual convention $\partial (b\otimes c)=(\partial b)\otimes c + (-1)^{|b|}b\otimes \partial c$

There is an inner-hom in $\mathbf{Chain}$, and the 1-chains are chain homotopies. The definition is $$ \underline{\mathbf{Chain}}(C_\bullet,D_\bullet)_k = \Pi_n \textrm{Hom} (C_{n-k},D_n) $$ so a 0-chain is just a map $f_n:C_n\rightarrow D_n$. The differential of this complex is given by $$ df(c) = d_D(f(c)) - (-1)^{|f|}\left(f(d_C(c)\right) $$ So a 0-cycle is just a chain map. A 1-chain whose boundary is $f-g$ is exactly a chain homotopy from $g$ to $f$.

Alternatively, there is a model structure on $\mathbf{Chain}$ where the weak equivalences are quasi-isomorphisms, and you can make sense of cylinder as a cylinder object for a chain complex, and then your topological motivation should all still make sense. I don't know all the details though so I won't try...

EDIT: crosspost...

This is mostly just to expand a little on John Klein and Alan Wilder's statements.

If you take the category of chain complexes and formally degree that quasi-isomorphisms $C \to D$ should become isomorphisms, you are already forced to identify chain-homotopic maps together.

Let $I$ be the chain complex which is $\mathbb{Z} \cdot s \oplus \mathbb{Z} \cdot t$ in degree 0, and $\mathbb{Z} \cdot H$ in degree 1, with $\partial H = t - s$. Then for any complex $C$, there are quasi-isomorphisms $i_s(c) = s \otimes c$ and $i_t(c) = t \otimes c$ from $C$ to $I \otimes C$, and an inverse quasi-isomorphism $p: I \otimes C \to C$ which kills $H$ and sends $s,t$ to $1$. We have $p i_s = id = p i_t$, and so if we turn quasi-isomorphisms into isomorphisms we get the identification $i_s = p^{-1} = i_t$.

A chain homotopy $H$ between two maps $f,g:C \to D$ is the same as a map $h: I \otimes C \to D$ such that $h i_s = f$, $h i_t = g$, with $h(H \otimes c) = H(c)$. If we have decreed quasi-isomorphisms to be isomorphisms, then we find $f = h i_s = h i_t = g$.

The miracle is that, for nice complexes, chain homotopy equivalence is enough. This is where the analogy with topological spaces comes in - the complex $I \otimes C$ is a cylinder object in a suitable model structure.