For the first question, the assomption is still true if you ask $G$ to be a discret group (or a Lie group) acting smoothly, freely and properly discontinuously on $M$.
This is a technical verification, if you need details this is really well treated in this book : Introduction to Smooth manifolds - J.M. Lee p153
For question 2), there are two points :
- the covering map $p:M\rightarrow M/G$ induces an injective map (the pull-back) $p^\star:H_{dR}^\bullet(M/G)\rightarrow H_{dR}^\bullet(M)$ :
Since $G$ is finite and $g^\star \circ p^\star=p^\star$ for every $g\in G$, if $p^\star \alpha=d\beta$ then for every $g\in G$, $g^\star \alpha = d\beta$ also. Hence you can write $p^\star \alpha= \dfrac{1}{\# G}\sum_{g\in G}g^\star d\beta=d(\dfrac{1}{\# G}\sum_{g\in G}g^\star \beta)=d\tilde \beta$. But $\tilde \beta$ is invariant under $G$ and $p^\star:\Omega^\bullet(M)\rightarrow \Omega^\bullet (M/G)$ is injective so $\alpha$ is exact.
- the image of this map is exactly $H_{dR}^\bullet(M)^G$.
For the third question, you can consider the orientable double cover $M^\star\rightarrow M=M^\star/G$ with $G=\mathbb Z/2\mathbb Z$. This is an orientable smooth manifold by construction and $M$ is not orientable iff $M^\star$ is connected.
$\triangleright$ That allows you to use the isomorphism $H_{dR}^n(M^\star)\cong \mathbb R$ given by integration on $M^\star$.
According to point 2), you have to show that $H_{dR}^n(M^\star)^G=\{0\}$ :
Let $\omega\in \Omega^n(M^\star)$ be a closed $n$-form. Note that the only non trivial deck transformation of the orientable double cover is the one that reverses the orientation on $M^\star$, let's denote this map by $f:M^\star\rightarrow M^\star$.
The closed $n$-form $\omega$ is invariant under the action of $G$ iff $f^\star \omega=\omega$ and under this assumption, we get : $\int_{M^\star}\omega=\int_{M^\star}f^\star\omega = -\int_{M^\star} \omega$ Hence $\omega$ is an exact n-form and $[\omega]=0\in H_{dR}^n(M^\star)$.