Let $I$ be a bounded below complex of injectives in some Abelian category, and $Z$ any bounded below complex. Suppose $u:I\to Z$ is a quasi isomorphism. We want to proof that $u$ is split injective up to homotopy. The prove outlined in Weibel's book An Introduction to Homological Algebra (pdf) is as follows:
The complex $\operatorname{cone}(u)$ is exact, where $$ \operatorname{cone}(u)^i=I^{i+1}\oplus Z^i, $$ and the differential is given by $$ D=\begin{pmatrix}-d & 0 \\ -u & d\end{pmatrix}. $$ The map $\phi:\operatorname{cone}(u)\to I[1]: (y,z)\mapsto -y$ is chain map, and it can be shown, using the fact that $I$ consists of injective objects, and $\operatorname{cone}(u)$ is exact, that $\phi$ is nullhomotopic, i.e. $$\phi = fD+df$$ for some $\{f^i\}$, where $$f^i:I^{i+1}\oplus Z^i\to I^i.$$ then we can write $f^i=(v^i,s^i)$, where $v^i:I^{i+1}\to I^i$ and $s^i:Z^i\to I^i$.
Then we get: \begin{align*} -y=\phi(y,z)&=fD(y,z)+df(y,z)\\ &=f(-dy,-uy+dz)+d(vy+sz)\\ &=-vdy-suy+sdz+dvy+dsz\\ &=-suy+(dvy-vdy)+(sdz+dsz). \end{align*} On the other hand, Weibel claims in his proof (after correcting a typo as per the errata) that $$y=(kdy+suy+dky)+(dsz-sdz).$$ Obviously this is not the same, and my answer does lead to a proof of the lemma, so I must be doing something wrong, but I don't see what.
Notice that you are looking at $I[1]$ instead of $I$, so there is an extra minus sign involving in the differential i.e. $d_{I[1]}f=-df$.