Remark: That's probably
To my opinion Benson's proof is correct and the same solution as that from mt. commutative diagramm in question ("drawing" the diagramms costed me a lot of time and a big disadvantage of mathoverflow is that it regularly happens that two persons are working second on a solution at the same time. Nevertheless, I post it, since I believe that the diagramms help understanding page 2 in the proof.pdf) always exists.
Consider the following commutative diagramm (it's the same as in your pdf but with named arrows):
$$P_2\quad \xrightarrow{D_2} \quad P_1 \quad \xrightarrow{D_1}\quad P_0 $$ $$ \hspace{2pt} \psi \downarrow \hspace{17pt} f_1 \downarrow \hspace{23pt} \downarrow f_2$$ $$N \quad \xrightarrow[d_2]{} \quad Y_1 \quad \xrightarrow[d_1]{} \quad Y_0$$
Futhermore let $h: P_1 \to N$ such that $\psi-\phi=h\circ D_2$, let $Z := P_1/D_2(\text{Ker } \phi)$ and let $\hat{f}_1 := f_1 -d_2\circ h: P_1 \to Y_1$.
Then on $\text{Ker }\phi$ holds: $$\hat{f}_1\circ D_2 = f_1D_2 -d_2hD_2 = f_1D_2 - d_2\psi = 0.$$ Thus $\hat{f}_1$ defines a map $\bar{f}_1: Z \to Y_1$ such that the right square in the following diagramm commutes:
$$N\quad \xrightarrow{i} \quad Z \quad \xrightarrow{D_1}\quad P_0 $$ $$ \hspace{35pt} id \downarrow \hspace{17pt} \bar{f}_1 \downarrow \hspace{25pt} \downarrow f_2\hspace{30pt}(\ast)$$ $$N \quad \xrightarrow[d_2]{} \quad Y_1 \quad \xrightarrow[d_1]{} \quad Y_0$$
Note that $i: N = \text{im }\phi \to Z$ is given by $i(n)=D_2(x) + \text{Ker } \phi$, if $x \in P_2$ such that $\phi(x)=n$. Hence the left-hand square also commutes.
But $(\ast)$ was just the commutative diagramm you were looking for.
