Let $G$ be a finite group acting freely on a manifold $M$, then we know that $\pi: M \to M/G$ is a covering map for manifolds. Now if I have a differential $k$-form $\omega$ on $M$, such that $g^{\ast}w = w$ for all $g \in G$. Why can I find a form $\bar w $ on $M/G$ such that $\pi^{\ast} \bar w = w$?
To get started. I can define $\bar w$ locally by $\bar w_y= (\pi_{x}^{-1})^{\ast}w_x$, where $x$ is in the fiber of $\pi$ over $y$. Is this the correct approach and how to show that it is well-defined? I assume it has some thing to do with $\omega$ being $G-$invariant.
I'm not sure what you mean by $\pi_x^{-1}$, but $\pi$ is not invertible.
Instead, for $y \in M/G$ and $Y_1, \dots, Y_k \in T_y(M/G)$ define
$$\overline{w}_y(Y_1, \dots, Y_k) := w_x(X_1, \dots, X_k)$$ where $x \in M$ and $X_i \in T_xM$ are such that $\pi(x) = y$ and $(\pi_*)_xX_i = Y_i$; note, as $\pi$ is a covering map, $(\pi_*)_x : T_xM \to T_y(M/G)$ is an isomorphism so $X_1, \dots, X_k$ exist and are unique.
The choice of the point $x$ is not unique, so we need to check that the form $\overline{w}$ is well-defined. In order to do this, suppose that $\pi(x') = y$ and $(\pi_*)_{x'}X'_i = Y_i$. As $\pi(x) = \pi(x')$, there is $g \in G$ such that $g(x) = x'$; here I am identifying $G$ with its image under the map $G \to \operatorname{Diff}(M)$ given by the group action. As $\pi = \pi \circ g$, we have $(\pi_*)_x = (\pi_*)_{g(x)}\circ (g_*)_x = (\pi_*)_{x'}\circ (g_*)_x$. Therefore $Y_i = (\pi_*)_x(X_i) = (\pi_*)_{x'}((g_*)_xX_i)$, so by the injectivity of $(\pi_*)_{x'}$, we see that $(g_*)_x(X_i) = X_i'$. So we see that
$$w_{x'}(X_1', \dots, X_k') = w_{g(x)}((g_*)_xX_1, \dots, (g_*)_xX_k) = (g^*w)_x(X_1, \dots, X_k) = w_x(X_1, \dots, X_k).$$
Therefore, the form $\overline{w}$ is well-defined.