EDITED, because I think I see the big picture now.
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 definition of homotopy is that $f=du+ud$. So my answer to your question is: In practice, if a map induces the zero map on homology, it is probably zero in the derived category. There do exist maps which are of the form $du+vd$, but are not of the form $du+ud$, see my other answer.
Paragraph about hereditary algebras deleted because I don't think it was quite right.
Some more elementary 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. $I$ is the chain complex of the obvious triangulation of the unit interval.
Thinking of $I$ as the unit interval, this really is a homotopy between $f$ and $0$.
I can't think of an analogous geometric motivation for $f=du+vd$.
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