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 "Introduction to compact transformation groups" and Proposition 1.1 of my note "The action on homology of finite groups of automorphisms of surfaces and graphs" available on my webpage <a href="http://www.math.rice.edu/~andyp/notes">here</a>.  Both of these are really about simplicial complexes and not manifolds, but it is standard that $M$ can be endowed with a triangulation that is preserved by $G$ (e.g. you can lift a triangulation of $M/G$ such that the images of all fixed points of $G$ are vertices).

If the characteristic of $F$ is not $0$, then all bets are off and in general you can't say anything.