Geometric Interpretation of Chain Homotpy Let $X$ and $Y$ be topological spaces. Two maps $f,g:X\to Y$ are said to be chain homotopic if for each $n$ we have a map $T_n:C_n(X)\to C_{n+1}(Y)$ such that $f_\sharp-g_\sharp=\partial_{n+1}T_n+T_{n-1}\partial_n$.
The notion of chain homotopy is used in proving that if the maps $f$ and $g$ are homotopic, then $f_*$ and $g_*$ behave exactly the same way between homology groups of $X$ and $Y$.
To do this, what the Algebraic Topology texts (like Rotman's Introduction to Algebraic Topology) do is construct a chain homotopy between $f_\sharp$ and $g_\sharp$.
Though the proof is neat, I do not see the intuition behind doing this. Is the notion of chain homotopy just a clever algebraic construct which makes life easy? Or is there a (strong?) geometric motivation behind it?
I have verified the details of the proof but if I were to reproduce the proof all I'll be doing is invoking things from memory with no insights of my own.
 A: Chain homotopies are an abstract way to define homotopy in the category $\mathsf{Ch}_\bullet$ of chain complexes. But indeed, there is a geometric interpretation of this.
$f, g : X \to Y$ be two homotopic maps, with homotopy given by $F : X \times [0, 1] \to Y$. $f_{\#}, g_{\#}$ be the induced maps $C_\bullet(X) \to C_\bullet(Y)$ between the singular chain complexes of $X$ and $Y$. We want to see what the correct analogue of the homotopy $F$ is in the context of chain complexes.
Well, let $\sigma : \Delta^n \to X$ be a singular simplex in $X$. Then $F \circ (\sigma \times \text{id}) : \Delta^n \times I \to Y$ is a singular prism inside $Y$ whose top $\Delta^n \times \{1\} \to Y$ corresponds to $f \circ \sigma$ and bottom $\Delta^n \times \{0\} \to Y$ corresponds to $g \circ \sigma$. This is already starting to look a bit like homotopies in the usual topological sense, where you use cylinders $X \times [0, 1]$ whose top is sent to $Y$ by $f$ and bottom is sent to $Y$ by $g$.
But prisms don't make sense in singular chain complexes, where chains are built by simplices. To make a chain out of this, we need to subdivide the singular prism into $(n+1)$-simplices. Hatcher explicitly describes the procedure in his book in page $112$. Here's the picture of the subdivision done for $\Delta^2 \times [0, 1]$ :

Once this is done, you can sum these $(n+1)$-simplices (with appropriate signs in their place - you have to take care of the orientations) to form a chain in $Y$.
Thus, you used $F$ to start from a singular $n$-simplex in $X$ and ended up with a singular $(n+1)$-chain in $Y$. This map $P_n : C_n(X) \to C_{n+1}(Y)$ is called a prism operator. This is precisely the analog for homotopies in $\mathsf{Top}$. Now, note that the boundary $\partial(\Delta^n \times [0, 1])$ of the prism is precisely sum of the top and bottom and sum of the sides $\partial(\Delta^n) \times [0, 1]$. Translating that into algebraic language and taking care of signs, we have
$$\partial_{n+1} P_n(\sigma) = f_{\#} - g_{\#} - P_{n-1}(\partial_n \sigma)$$
Rearranging this, we get that formula you mentioned :
$$\partial_{n+1} P_n + P_{n-1} \partial_n = f_{\#} - g_{\#}$$
This is in general used as a notion of homotopy of abstract chain complexes in homological algebra, as it matches with the topological notion of homotopy in the context of singular chain complexes as we saw above.
