= Solution
First work at the level of <differential forms>. The fiber description implies $\pi\circ f=\pi$, so every pullback $\pi^*\beta$ is $f$-invariant. Since $\pi$ is a <surjective> <local diffeomorphism>, its pullback on forms is <injective>: at any $y\in N$, choose a lift $x$ and use the inverse of $d\pi_x$ to evaluate the form on arbitrary tangent vectors at $y$.
Conversely let $\alpha$ be an $f$-invariant form on $M$. For a local inverse $s:V\to M$ of $\pi$, define $\beta|_V=s^*\alpha$. If two inverses select the same lift they agree locally; if they select different lifts, the fiber condition identifies the second with $f$ composed with the first. Locally these alternatives are stable, using disjoint inverse neighborhoods at distinct lifts. The equality $f^*\alpha=\alpha$ therefore makes the local definitions agree. They glue to a unique smooth form $\beta$ on $N$ with $\pi^*\beta=\alpha$. Thus
$$
\pi^*:\Omega^p(N)\cong\Omega^p(M)^f
$$
is an isomorphism of complexes, with the <exterior derivative> on both sides.
It remains to compare the <cohomology> of the invariant complex with invariant <de Rham cohomology> classes. Put $P=(I+f^*)/2$ on forms. If $f^*[\alpha]=[\alpha]$ and $d\alpha=0$, then $P\alpha$ is invariant, closed and represents the same class, because its class is the average of two equal classes. If an invariant form is exact, $\alpha=d\eta$, then $\alpha=d(P\eta)$ with an invariant primitive. This proves both surjectivity and injectivity on <cohomology>. For $p=0$ there are no negative-degree primitives, and the same conclusion follows directly from closed functions.
Combining the two steps gives
$$
\boxed{\pi^*:H^p_{\mathrm{dR}}(N)\xrightarrow{\ \cong\ }H^p_+(M).}
$$
This is the <de Rham cohomology of a finite quotient> argument specialized to an involution. It does not require <orientation>. The hypotheses do not explicitly exclude $f=\operatorname{id}$; then all fibers are singletons, $\pi$ is a bijective <local diffeomorphism> and the conclusion is the ordinary <diffeomorphism> invariance. If $f$ has one fixed point, local injectivity of $\pi$ makes its fixed set open as well as closed; connectedness of $M$ forces this identity case. Otherwise the fibers have two points and $\pi$ is the usual double covering.
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