Here is a version of the proof that avoids spectral sequences by spelling out all the definitions:
Define the mapping torus $T(M, \phi) := (M\times [0,1])/(x, 0)\sim (\phi(x),1)$. The projection to the first factor defines a map $p: T(M, f)\to M/\mathbb Z$, which isinduces an isomorphism on cohomology since the action is free and proper (cover $M$ with open sets $U_i$ which are equivariantly homeomorphic to $(U_i/\mathbb Z)\times Z$; the condition is easy to check for each $U_i$ and follows by induction for finite unions of $U_i$'s from the five-lemma applied to the Mayer-Vietoris sequence; for infinite unions, apply the five-lemma to the Milnor exact sequence). The pair $(T(M, \phi),[M\times\{0\} ])$ then gives a long exact sequence in cohomology. Since $[M\times\{0\}]\subset T(M,\phi)$ and $M\times\{0, 1\}\subset M\times [0,1]$ have the homotopy extension property, the quotient map induces an isomorphism $H^n(M\times [0, 1],M\times \{0,1\})\cong H^n(T(M, \phi), [M\times\{0\}])$. Since the inclusion $M\times \{0,1\}\subset M\times [0,1]$ induces the diagonal on cohomology, the long exact sequence of cohomology groups provides an isomorphism $\big(H^n(M)\oplus H^n(M)\big)/\Delta\cong H^{n+1}(M\times [0, 1],M\times \{0,1\})$. The source is of course isomorphic to $H^n(M)$ via $(x, y) \mapsto x - y$. Since the composition $M\times \{1\}\to [M\times \{1\}] = [M\times \{0\}] \cong M\times \{0\}$ is given by $f$, the boundary operator of the pair $(T(M, \phi), [M\times \{0\}])$ can be identified with the map $\operatorname{id} - f^*: H^n(M)\to H^n(M)$, and consequently there are short exact sequences
$$ 0\to \operatorname{coker}(H^{n-1}(M)\xrightarrow{\operatorname{id} - f^*} H^{n-1}(M)\big)\to H^n(T(M,\phi))\to \operatorname{ker}(H^{n}(M)\xrightarrow{\operatorname{id} - f^*} H^{n}(M)\big)\to 0 $$
The target are precisely the invariants $H^n(M)^\mathbb Z$.