Why chain homotopy when there is no topology in the background? Given two morphisms between chain complexes  $f_\bullet,g_\bullet\colon\,C_\bullet\longrightarrow D_\bullet$, a chain homotopy between them is a sequence of maps $\psi_n\colon\,C_n\longrightarrow D_{n+1}$ such that $f_n-g_n= \partial_D \psi_n+\psi_{n-1}\partial_C$. I can motivate this definition only when the chain complexes are associated to some topological space. For example if $C_\bullet$ and $D_\bullet$ are simplicial chain complexes, then for a simplex $\sigma$, a homotopy between $f(\sigma)$ and $g(\sigma)$ is something like $\psi(\sigma)\approx\sigma\times [0,1]$, whose boundary is $f(\sigma)-g(\sigma)-\psi(\partial\sigma)$.
But chain homotopy also features in contexts where there is no topological space lurking in the background. Examples are chain homotopy of complexes of graphs (e.g. Conant-Schneiderman-Teichner) or chain homotopy of complexes in Khovanov homology. In such contexts, the motivation outlined in the previous paragraph makes no sense, with "the boundary of a cylinder being the top, bottom, and sides", because there's no such thing as a "cylinder". Thus, surely, the "cylinder motivation" isn't the most fundamental reason that chain homotopy is "the right" relation to study on chain complexes. It's an embarassing question, but:

What is the fundamental (algebraic) reason that chain homotopy is relevant when studying chain complexes?

 A: Here is how I think about chain complexes.  (I mention that you should always allow yourself complexes that go in both directions.)
There is a category of $\mathbb Z$-graded vector spaces, which is monoidal (just because it's the category of $\mathbb G_m$-modules), but I give it the interesting symmetric structure following the usual "super" sign rules --- let $\mathfrak Q$ denote the one-dimensional vector space in degree $1$; then I choose the braiding $\mathfrak Q \otimes \mathfrak Q \to \mathfrak Q \otimes \mathfrak Q$ to be minus identity (this determines the braiding on the category).
Now, give the object $\mathfrak Q$ the structure of an abelian Lie algebra.  (Note that by "Lie algebra" and "representation" I mean that you should interpret all AS/IHX/STU with respect to the chosen braiding.)
Then the category of Chain Complexes, as a symmetric monoidal category, is precisely the category of (left, say) $\mathfrak Q$-modules: checking this is a nice exercise (and it explains what are the correct signs).
But the category of $\mathfrak Q$-modules, as Alan Wilder has said, has an inner hom, and you can tell what it is because it's an inner hom of representations of a Lie algebra.  Namely, if $f\in \underline{\hom}(X,Y)$, then the basis vector $d\in \mathfrak Q$ acts on $f$ by $f \mapsto [d,f]$, where the commutator is, of course, to be interpreted in the super sense, which is the sense internal to the category.
So chain maps are the invariant global (= degree-0) elements.  (degree-(-1)) $h$ is a homotopy between (degree-0) $f,g$ iff $[d,h] = f-g$, as Tom pointed out.  Another way to say this: $f-g$ is in the $\mathfrak Q$-orbit generated by the degree-(-1) elements, so that $f,g$ are in the same coset thereof.
A: An important application of chain homotopy is in (co)homology: If $f,g: C \to D$  are homotopic then $$[f_n] = [g_n]: H_n(C) \to H_n(D).$$
Moreover chain homotopy has good functorial properties. For example, with $f,g$, the chain maps 
$$ Hom(f,id_B), Hom(g,id_B): Hom(D,B) \to Hom(C,B)$$ 
are again homotopic, implying
$$ [Hom(f_n,id_B)] = [Hom(g_n,id_B)]: H^n(D,B) \to H^n(C,B).$$ 
As an application, one can show that induced homomorphisms for $Ext$ are unique. For: Let $C \to A$ resp. $D \to A'$ be a projective resolution of $A$ resp. $A'$ and let $\alpha: A \to A'$ be a homomorphism. By the fundamental lemma of projective resolutions, two chain maps $f, g$ that extend $\alpha$ are homotopic, so by the above, 
$$ \alpha^*_n := [Hom(f_n,id_B)]: Ext^n(A',B) \to Ext^n(A,B)$$ is well-defined, where $Ext^n(A,B):= H^n(C,B)$.
A: This was a comment, but I guess it was not so clear, although it is hiding in some of the above answers:
There is an object in $Ch^+(R)$ that behaves like an interval. Think of the simplicial interval, and you get a chain complex $I_*$ over $R$ with $I_0=Re_0 \oplus Re_1$, $I_1=Rf_0$ and all other groups 0, with differential $d(f_0)=e_1−e_0$. This plays the role of the unit interval. Now a chain homotopy of two chain maps $f,g:A_∗ \to B_∗$ really is $H:A_∗ \otimes I_* \to B_*$ where the tensor product takes place in R chain complexes. I am not sure how well this object behaves with respect to the Dold-Kan maps, but it certainly is the normalized chains on $\Delta^1$. You could similarly look at maps of $R$ complexes from $I_*$ into $Hom_R(A_*,B_*)$.
With this in hand you can form cylinder objects and cofiber sequences in the "obvious" way, and things start looking a bit more geometric.
A: 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 $\partial (b\otimes c)=(\partial b)\otimes c + (-1)^{|b|}b\otimes \partial c$
A: 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...
A: 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.
