Despite somewhat rusty homology knowledge, I think I can follow the proof.
1) Just to repeat my comment on this, there's actually not much too it. If $\tau$ is a simplex in $Y$ and $\sigma$ is a lifting of $\tau$ to $X$, then $g\circ\sigma$ is another lifting of $\tau$.
2) We need some slight prior knowledge: for any $p>0$, $ H_p(S^p,\mathbb{Z}_2)\approx H_0(S^p,\mathbb{Z}_2)\approx\mathbb{Z}_2 \textrm{ while } H_q(S^p,\mathbb{Z}_2)\approx0\textrm{ for }q=1,\ldots,p-1. $
2a) The first step is to prove that $H_q(P^p,\mathbb{Z}_2)\approx\mathbb{Z}_2$ for $q=0,\ldots,p$. To do this, we use the long exact sequence $ \cdots \rightarrow H_q(P^p,\mathbb{Z}_2) \rightarrow H_q(S^p,\mathbb{Z}_2) \rightarrow H_q(P^p,\mathbb{Z}_2) \rightarrow H_{q-1}(P^p,\mathbb{Z}_2) \rightarrow H_{q-1}(S^p,\mathbb{Z}_2) \rightarrow\cdots $ where $H_q(P^p,\mathbb{Z}_2)\rightarrow H_q(S^p,\mathbb{Z}_2)$ maps a simplex to the sum of both liftings of the simplex, while $H_q(S^p,\mathbb{Z}_2)\rightarrow H_q(P^p,\mathbb{Z}_2)$ just maps a simplex to the image in the ordinary way. If $2, we get $H_q(P^p,\mathbb{Z}_2)\approx H_{q-1}(P^p,\mathbb{Z}_2)$ since $H_q(S^p,\mathbb{Z}_2)\approx H_{q-1}(S^p,\mathbb{Z}_2)\approx0$.
At the lower end of the sequence we get $ 0 \rightarrow H_1(P^p,\mathbb{Z}_2) \rightarrow H_0(P^p,\mathbb{Z}_2) \rightarrow H_0(S^p,\mathbb{Z}_2) \rightarrow H_0(P^p,\mathbb{Z}_2) \rightarrow 0 $ where the first and the last maps are isomorphisms, and get $H_1(P^p,\mathbb{Z}_2)\approx H_0(P^p,\mathbb{Z}_2)\approx\mathbb{Z}_2$.
At the upper end of the sequence, we get $ 0\rightarrow H_p(P^p,\mathbb{Z}_2) \rightarrow H_p(S^p,\mathbb{Z}_2) \rightarrow H_p(P^p,\mathbb{Z}_2) \rightarrow H_{p-1}(P^p,\mathbb{Z}_2) \rightarrow 0 $ where the middle map must be zero, hence $H_p(P^p,\mathbb{Z}_2)\approx H_{p-1}(P^p,\mathbb{Z}_2)\approx\mathbb{Z}_2$. The reason the middle map is zero is that the composition $H_p(S^p,\mathbb{Z}_2)\rightarrow H_p(P^p,\mathbb{Z}_2)\rightarrow H_p(S^p,\mathbb{Z}_2)$, which are first two maps of the long sequence just composed in the opposite order, has to be zero (since it maps a simplex to twice itself); but $H_p(P^p,\mathbb{Z}_2)\rightarrow H_p(S^p,\mathbb{Z}_2)$ is injective (monomorphism), so for the composition to be zero the map $H_p(S^p,\mathbb{Z}_2)\rightarrow H_p(P^p,\mathbb{Z}_2)$ must be zero.
2b) The next step is to apply the assumed equivariant map $\phi:S^n\rightarrow S^m$ for $n>m$ to obtain a contradiction.
The map $\phi_*:H_k(P^n,\mathbb{Z}_2)\rightarrow H_k(P^m,\mathbb{Z}_2)$ is an isomorphism for $k=0,\ldots,m$. This follows e.g. by induction, starting with $k=0$ and using $H_k(P^n,\mathbb{Z}_2)\approx H_{k-1}(P^n,\mathbb{Z}_2)$ and $H_k(P^m,\mathbb{Z}_2)\approx H_{k-1}(P^m,\mathbb{Z}_2)$ (commutative diagram in the book).
The final commutative diagram is now $ \begin{array}{ccc} \mathbb{Z}_2\approx H_m(P^n,\mathbb{Z}_2)&\rightarrow &H_m(S^n,\mathbb{Z}_2)\approx 0\\ \downarrow&\circ&\downarrow\\ \mathbb{Z}_2\approx H_m(P^m,\mathbb{Z}_2)&\rightarrow &H_m(S^m,\mathbb{Z}_2)\approx\mathbb{Z}_2\\ \end{array} $ where going in one direction should produce an isomorphism, while going in the other should map to zero since $H_m(S^n,\mathbb{Z}_2)\approx 0$. I.e., we have a contradiction.