Let $(E,\partial)$ be a differential space. A linear map $h:E\to E$ is a homotopy operator if $h\partial+\partial h = {\rm Id}$.
Then, there is a homotopy operator in $E$ if and only if $H(E)=0$.
Checking that such an $h$ forces $H(E)=0$ isn't hard, but I'm having trouble constructing $h$ if $H(E)=0$. I'm happy with hints, since I'm doing this just for fun. This is an exercise in Greub's Linear Algebra book (I don't recall which exercise exactly, but the book is awesome!).
As you said, one way is trivial. For the other, split $E=E' \oplus F$, where $E'=\ker \partial= \text{Im}~ \partial$ and $F$ is some subspace (you use the fact that we are dealing with vector spaces here). Then define
$$h(e'+f)=\partial|_{F}^{-1}e'.$$
Since $\partial|_{F}: F \to E'$ is a bijection (why?), $h$ is well-defined. Note now that $$\partial h (e'+f)+h \partial(e'+f)=\partial(\partial|_{F}^{-1} e')+h(\partial f)=e'+f.$$
OBS: Caveat lector. You have a very particular chain complex here (a "constant" one), but a chain homotopy raises degree in general. Be careful not to be confused with this (and a lot of quite particular instances in this exercise) in the future when you encounter actual chain complexes in homology.