Conjugating $\Gamma(N)$ by $\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix}$
$$\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix} \displaystyle\Gamma(N)\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & 1/N \end{bmatrix}
\scriptstyle \ \ = \ \ \underbrace{\left\{\ \scriptstyle\begin{bmatrix} aN+1 & b \\ cN^2 & dN+1\end{bmatrix} \in SL_2(\mathbb{Z})\right\}}_{\displaystyle\tilde{\Gamma}_1(N^2)}\displaystyle \ \supset \Gamma_1(N^2)$$

Define the linear operator $T : M_k(\Gamma(N)) \to M_k(\tilde{\Gamma}_1(N^2)), \ \ T f(\tau) = f (N\tau)$,

and the usual linear operators for showing $M_k(\Gamma_1(N^2)) =\displaystyle \bigoplus_{\chi \bmod N^2}M_k(\Gamma_0(N^2),\chi)$

For $gcd(d,N^2)=1$, let $\langle d \rangle : M_k(\Gamma_1(N^2)) \to M_k(\Gamma_1(N^2)) $, $\ \ \langle d \rangle g = g|_k\gamma, \quad \gamma \in \Gamma_0(N^2), \quad\gamma_d \equiv d \bmod N^2$ (which is well-defined, not depending on the chosen $\gamma$). Note that $\langle d d' \rangle = \langle d \rangle\langle d' \rangle$

And for a $\chi \bmod N^2$ :
$$ \pi_\chi g= \frac{1}{\varphi(N^2)}\sum_{\begin{array}{l}d \bmod N^2\\gcd(d,N^2) =1\end{array}} \overline{\chi(d) }\langle d \rangle g$$
$\pi_\chi$ is an orthogonal projection $M_k(\Gamma_1(N^2)) \to M_k(\Gamma_0(N^2),\chi)$, and $\displaystyle\sum_{\chi \bmod N^2} \pi_\chi g=\langle 1\rangle g= g$ and for any $\chi \ne \chi'$ : $\pi_\chi \pi_{\chi'} = 0$

Finally, $Tf = \langle dN+1\rangle T f$, so that $\langle dN+d'\rangle T f= \langle d'\rangle T f$ and hence $\pi_\chi T f = 0$ whenever $\chi$ isn't a character $\bmod N$.

Thus $\sum_{\chi \bmod N} \pi_\chi Tf = Tf$, and together with $M_k(\Gamma_0(N^2),\chi) \subset M_k(\tilde{\Gamma}_1(N^2))$ for any $\chi \bmod N$,

it means that $$M_k(\Gamma(N)) \simeq M_k(\tilde{\Gamma}_1(N^2)) =\bigoplus_{\chi \bmod N}M_k(\Gamma_0(N^2),\chi)$$

4more comments