Here are <strike>two</strike> three observations:

(1) It is required that $f$ be a map of chain complexes, so $df=fd$. So we want $d(du+vd) =(du+vd)d$ or $dud=dvd$. This doesn't force $u=v$, but it is the easiest way to achieve it.

(2) There is a topological way of thinking of the condition $du+ud=f$, which I learned from Joel Kamnitzer. Let $I$ be the chain complex with $I_1=\mathbb{Z}$, $I_0 = \mathbb{Z}^2$ and the map $I_1 \to I_0$ given by $(1 \ -1)$. Let $\partial I$ be the subchain complex where $(\partial I)\_0=I_0$ and $(\partial I)\_i=0$ for all other $i$.

Then writing $f=du+ud$ is equivalent to finding a map $u:C \times I \to D$ such that, when we restrict to $C \times (\partial I)$, we have the map $f$ on one component and $0$ on the other. Thinking of $I$ as the unit interval, this really is a homotopy between $f$ and $0$. (And $I$ is the chain complex of the obvious triangulation of the unit interval.) 

I can't think of an analogous geometric motivation for $f=du+vd$.

(3) We have the following theorem: let $C$ and $D$ be complexes of projective objects. The following are equivalent: (a) the map $f: C \to D$ is the zero map in the derived category (b) there is a homotopy between $f$ and $0$.

In this theorem, the homotopy is using the definition $du+ud$. <strike>I'm not sure what happens if you consider homotopies of the form $du+vd$. I can imagine two answers: maybe these maps are not zero in the derived category, or maybe they are zero and the above theorem says that any maps which can be realized as $du+vd$ can also be realized as $du+ud$.</strike> I now have an example of a map which can be realized as $du+vd$ but not as $du+ud$; see my other answer.