Let $\Gamma$ be a congruence subgroup which contains some level $n$ subgroup $\Gamma(n)$ and $f$ be a meromorphic modular form(i.e. for all $L\in SL_2(Z),\exists C_L>0, f|L^{-1}=f(L^{-1}z)\sqrt{cz+d}^r$ with $c,d$ integer entries from $L$, is holomorphic above $Im(z)>C_L$ and they do not have essential singularity at $i\infty$) of weight $\frac{r}{2}$ with multiplier system $v:\Gamma\to C^\star$(This does not have to be a group homomorphism but each $v(G)$ is some roots of unity for all $G\in\Gamma$.) $f$ satisfies $f|L=v(L)f$ for $L\in\Gamma$. $f|L^{-1},L\in SL_2(Z)$ has a different multiplier system induced by $L$ conjugation on $\Gamma$ s.t. $G\in L\Gamma L^{-1},(f|L^{-1})|G=\tilde{v}(G)f|L^{-1}$.
Since $\Gamma$ contains some level $n$ subgroup, clearly there is a translation matrix $G$ acting upon $f$ s.t. $f(z+R)=v(G)f(z)$. Say the translation is exactly by $R$ units along $x$ direction and $v(G)^l=1$, the book claims that $f(z+Rl)=f(z)$.
$\textbf{Q1}$: Why $f(z+Rl)=f(z)$ holds? I cannot see this hold in general setting as I can do $f(z+2Rl)=f(z)$ at best. Consider $(f|G^l)(z)=v(G^l)f(z)$. Now $A,B\in\Gamma,v(AB)=w(A,B)v(A)v(B)$ where $w(A,B)$ is $\pm 1$. I have checked that $w(A,I)=1,v(I)=1$ for $I$ being identity matrix. So in total, I missed a sign if I use plainly $f(z+Rl)=f(z)$.
Since $f$ is periodic along $N=Rl$, I can perform fourier expansion/q-expansion $f=\sum_{n\in Z}a_nq^n$ where $q=e^{\frac{2\pi i z}{R}}$. Say $v(G)=e^{\frac{2\pi iv}{l}},(v,l)=1$. From relation $f(z+R)=v(G)f(z)$, I can obtain relation $a_{v+kl}$ only non-trivial coefficients. So I obtain $f=\sum_{k\in Z}a_{v+kl}q^{v+kl}$. Since $f$ is meromorphic, it cannot have essential singularity at $i\infty$. This fourier expansion must start from some finite negative $k$ value at worst. The book claim that $f(i\infty)=0$.
$\textbf{Q2:}$ Why $f(i\infty)=0$? It is very possible that $q$ expansion has non-trivial pole.
$Ord_\Gamma(f,i\infty)=min\{v+kl\vert a_{v+kl}\neq 0\}$
Let $N\in SL_2(Z)$ be a matrix with $N(i\infty)=i\infty$. Then $Ord_\Gamma(f,i\infty)=ord_{N\Gamma N^{-1}}(f|N^{-1},i\infty)$
$\textbf{Q3}:$ Is there an easy way to see $Ord_\Gamma(f,i\infty)=ord_{N\Gamma N^{-1}}(f|N^{-1},i\infty)$? I tried to compute the $q-$expansion but I do not see how to see the correspondence of $q-$expansion here or the transformation on $q-$expansion.