Let $M,N$ be positive integers and $\mathfrak{h}=\{x+iy:y>0, x\in \Bbb R\}\subseteq \Bbb C$. Suppose $f:\mathfrak{h}\to\Bbb C$ satisfies $f(\frac{a\tau+b}{c\tau+d})=(c\tau+d)^kf(\tau)$ for all $\tau\in\mathfrak{h}$, $\begin{pmatrix} a & b\\ c & d \end{pmatrix}\in \Gamma_0(M):=\left\{\begin{pmatrix} a & b\\ c & d \end{pmatrix}\in SL_2(\Bbb Z): M|c\right\}$. We define the function $f_1:\mathfrak{h}\to\Bbb C$ by $f_1(\tau):=f(N\tau)$. We want to show that $f_1\left(\frac{a\tau+b}{c\tau+d}\right)=(c\tau+d)^kf_1(\tau)$ for all $\tau\in\mathfrak{h}$, $\begin{pmatrix} a & b\\ c & d \end{pmatrix}\in \Gamma_0(MN)$.
Attempt. For any $\begin{pmatrix} a & b\\ c & d \end{pmatrix}\in\Gamma_0(MN)$ and $\tau\in\mathfrak{h}$, since $f_1\left(\frac{a\tau+b}{c\tau+d}\right)=f\left(\frac{Na\tau+Nb}{c\tau+d}\right)$ and $(c\tau+d)^kf_1(\tau)=(c\tau+d)^kf(N\tau)$, it is enough to show that $f\left(\frac{Na\tau+Nb}{c\tau+d}\right)=(c\tau+d)^kf(N\tau)$. I couldn't go further from here.
Let $\gamma=\begin{bmatrix}a & b\\ c & d\end{bmatrix}$ where $c=kMN$ then $$f_1(\gamma\tau)=f(N\gamma\tau)=f\left(\begin{bmatrix}Na & Nb\\ kMN & d\end{bmatrix}\tau\right)=f\left(\begin{bmatrix}a & Nb\\ kM & d\end{bmatrix}N\tau\right)$$ Since $ad-kMNb=ad-bc=1$ we have $$f\left(\begin{bmatrix}a & Nb\\ kM & d\end{bmatrix}N\tau\right)=(kMN\tau+d)^mf(N\tau)=(c\tau+d)^mf_1(\tau)$$ which completes the proof.
You might wanna check the analyticity of $f_1$ at its cusps if you are proving that it is a modular form. But if the identity is the ultimate goal then the previous thing should suffice. Also, I used $m$ instead of $k$ for weight.