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As far as I understand your question, you want to see a derivation of the formula for $\chi(M/G)$. Here it is:

1. The difficult part of the argument os to show that there is an isomorphism $H^* (M/G; \mathbb{Q}) \cong H^* (M; \mathbb{Q})^G$ (the $G$-invariants). It is induced from the quotient map $M \to M/G$, but that is not so important. In the following, all cohomology have rational coefficients.

2. If that is done, the argument is easy. By elementary representation theory of finite groups:

$$dim (H^i (M)^G) = \frac{1}{|G|} \sum_{g \in G} Tr_{H^i (M)} (g) g).$$

by elementary representation theory of finite groups.

1. We show the difficult part in two steps. Consider the Borel construction $EG \times_G M$. There is a fibre bundle $\pi:EG \times_G M \to BG$ with fibre $M$ and a map $f: EG \times_G M \to M/G$.

2. The Leray-Serre spectral sequence of $\pi$ begins with $E_{2}^{pq}=H^p (G;H^q (M))$. If $p > 0$, this group is zero and hence $H^n(EG \times_G M) \cong H^0 (G, H^n(M))=H^n (M)^G$.

3. $f$ induces an isomorphism in cohomology: Let $x \in M$ and let $H$ be the stabilizer subgroup at $x$. Pick a $G$-equivariant Riemann metric on $M$. Consider $V=\bigcup_{g \in G} B_{\epsilon}(x)$. If $\epsilon$ is small enough, then $V$ is a disc bundle a $G$-equivariant vector bundle on $G/H$. Clearly, $EG \times_G V \simeq EG \times_G G/H \simeq BH$. Moreover, $V/G\simeq \mathbb{R}/H$ is contractible. The consequence of this discussion is that there exists a finite cover of $M/G$ by open contractible sets, such that all intersections are again contractible or empty. Also, the preimages of the covering sets and their intersections have trivial rational cohomology, because the cohomology of $BH$ is trivial. By the Mayer-Vietoris sequence, induction on the number of covering sets and repeated application of the 5-lemma, it follows that $f^* :H^* (M/G) \to H^* (EG \times_G M)$ is an isomorphism in rational cohomology.

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As far as I understand your question, you want to see a derivation of the formula for $\chi(M/G)$. Here it is:

1. The difficult part of the argument os to show that there is an isomorphism $H^* (M/G; \mathbb{Q}) \cong H^* (M; \mathbb{Q})^G$ (the $G$-invariants). It is induced from the quotient map $M \to M/G$, but that is not so important. In the following, all cohomology have rational coefficients.

2. If that is done, the argument is easy:

$$dim (H^i (M)^G) = \frac{1}{|G|} \sum_{g \in G} Tr_{H^i (M)} (g)$$

by elementary representation theory of finite groups.

1. We show the difficult part in two steps. Consider the Borel construction $EG \times_G M$. There is a fibre bundle $\pi:EG \times_G M \to BG$ with fibre $M$ and a map $f: EG \times_G M \to M/G$.

2. The Leray-Serre spectral sequence of $\pi$ begins with $E_{2}^{pq}=H^p (G;H^q (M))$. If $p > 0$, this group is zero and hence $H^n(EG \times_G M) \cong H^0 (G, H^n(M))=H^n (M)^G$.

3. $f$ induces an isomorphism in cohomology: Let $x \in M$ and let $H$ be the stabilizer subgroup at $x$. Pick a $G$-equivariant Riemann metric on $M$. Consider $V=\bigcup_{g \in G} B_{\epsilon}(x)$. If $\epsilon$ is small enough, then $V$ is a disc bundle a $G$-equivariant vector bundle on $G/H$. Clearly, $EG \times_G V \simeq EG \times_G G/H \simeq BH$. Moreover, $V/G\simeq \mathbb{R}/H$ is contractible. The consequence of this discussion is that there exists a finite cover of $M/G$ by open contractible sets, such that all intersections are again contractible or empty. Also, the preimages of the covering sets and their intersections have trivial rational cohomology, because the cohomology of $BH$ is trivial. By the Mayer-Vietoris sequence, induction on the number of covering sets and repeated application of the 5-lemma, it follows that $f^* :H^* (M/G) \to H^* (EG \times_G M)$ is an isomorphism in rational cohomology.