If $\Gamma$ is of finite index in $SL_2(\mathbb{Z})$ , show that the number of $\Gamma$-inequivalent cusps is $\leq |SL_2(\mathbb{Z}:\Gamma)|$ .
Proof . Let $s \in \mathbb{Q}$ $\cup$ $ \lbrace i \infty \rbrace$ be a cusp . Then there exists $\tau$ $\in SL_2(\mathbb{Z})$ such that $s=\tau(i\infty)$ . If $g_1,...,g_r$ are right coset representatives of $\Gamma$ in $SL_2(\mathbb{Z})$ , then
$s=\tau(i\infty)=\gamma g_j(i\infty)$ for some $\gamma\in \Gamma$ . (1)
Hence the set of Γ-inequivalent cusps is contained in the set $$\lbrace g_1(i\infty),...,g_r(i\infty) \rbrace $$ whose size is at most |$SL_2(\mathbb{Z}):\Gamma$| .
I do not really understand what is meant by inequivalent cusps and what is meant by "...for some $\gamma \in \Gamma "$ . I think it does not mean that there exists only one $\gamma$ .
Does (1) mean that there exists $\gamma \in \Gamma$ such that $s=\gamma g_j(i\infty)$ for all j ? But then I think there is no $\Gamma$-inequivalent cusp .
If we care of cusps it is because for a finite index subgroup $\Gamma \le SL_2(Z)$ then $$\Gamma \setminus H = \{ \Gamma z, z \in H\}$$ is a Riemann surface
(where $\Gamma z = \{\gamma z, \gamma \in \Gamma\}$ is a subset of $H$, so each point of that Riemann surface is a subset of $H$, in the same way that each point of the complex torus $C/(Z+iZ)$ is a subset of $C$).
But it is a non-compact Riemann surface, so its field of meromorphic function is way too large.
While $$\Gamma \setminus H^*=\{ \Gamma z, z \in H \cup Q \cup i\infty\}$$ is a compact Riemann surface.
$Q \cup i\infty = SL_2(Z) i \infty$ thus the added points are of the form $\Gamma \alpha i\infty$ with $\alpha \in SL_2(Z)$, thus $\Gamma \alpha \in \Gamma \setminus SL_2(Z)$ (this notation means quotient on the left) and the size of $\Gamma \setminus SL_2(Z)$ is $[SL_2(Z):\Gamma]$.
Two rational numbers are "equivalent cusps" iff $\Gamma u= \Gamma v$.
That the obtained Riemann surface is compact is seen in the fact it is covered by finitely many charts from the closed unit disk.
About the modular forms : $f \in M_{12}(\Gamma)$ iff $\frac{f(z)}{\Delta(z)}$ is meromorphic on $\Gamma \setminus H^*$ with no poles on $\Gamma \setminus H$ and at most simple poles at the cusps. This is exactly what Riemann-Roch is about : the meromorphic functions on a compact Riemann surface with prescribed poles. We look at $f$ instead of $\frac{f}{\Delta}$ because the Hecke operators act in a simple way on them and their eigenvalues correspond to the Fourier coefficients.