Choose a $G$-invariant Riemannian metric on $M$. For any $x\in M$ let $U_x$ be the tangent space to the orbit $Gx$ at $x$, and let $V_x$ denote the orthogonal complement of $U_x$ in $T_xM$. As the action is free, $U_x$ is canonically identified with the Lie algebra $LG$. The spaces $V_x$ give a subbundle of $TM$. Let $p:M\to M/G$ be the projection. For $y\in M/G$, let $W_y$ be the set of equivariant sections of $V|{p^{-1}\{y\}}$. As $p^{-1}\{y\}$ is a free $G$-orbit, we have natural identifications $W_y\simeq V_x$ for all $x\in p^{-1}\{y\}$. It is also not hard to see that $W$ is the tangent bundle for $M/G$. (You have to do most of this work to make $M/G$ a manifold in the first place.) Put $n=\dim(M)$ and $d=\dim(G)$. It now follows that sections of $\Lambda^{n-d}(T(M/G))$ biject with $G$-invariant sections of $\Lambda^{n-d}(V)$. If we pick a nonzero element $u$ in the space $\Lambda^d(LG)\simeq\mathbb{R}$ then multiplication by $u$ gives a $G$-equivariant isomorphism
$$ \Lambda^{n-d}(V) \to \Lambda^n(LG\oplus V) = \Lambda^n(TM). $$
Thus, orientations of $M/G$ biject with $G$-invariant orientations of $M$ (but the bijection depends on the sign of $u$, or in other words the orientation of $LG$).