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This is true even if the group does not act freely. See Proposition 1.1 of my notes here. I deal with simplicial complexes and work over the rationals, but the statement you give can be proved the s …

If $F$ is a field of characteristic $0$, then $H^k(M/G;F)$ equals the invariants of the action of $G$ on $H^k(M;F)$. For two different proofs of this, see Proposition III.2.4 of Bredon's "Introductio …

If $G$ is a finite group acting by simplicial automorphisms on a simplicial complex $X$, then $$H^{\ast}(X/G;\mathbb{Q}) = (H^{\ast}(X;\mathbb{Q}))^{G} = H^{\ast}((C^{\ast}(X;\mathbb{Q}))^G),$$ whe …

It's terrible notation to use $p$ for both the prime in question and for the index in the spectral sequence, so I'll write the spectral sequence as $$E_2^{nm} = H^n(G;H^m(X)).$$ Since $X$ is a mod-$p$ …

Since you also asked about alternate proofs of Theorem 1, there is a proof that does not use the localization theorem of a stronger version of the inequality (due to Floyd) in my notes Smith theory an …