I know that $M_2(\Gamma_0(4))$ is generated by
$E_{2,2}(z)=E_2(z)-2E_2(2z)$ and
$E_{2,4}(z)=E_2(z)-4E_2(4z)$,
where $E_2$ is the Eisenstein's series.
This space is supposed to only have $0$ as cusp form (i.e. $S_2(\Gamma_0(4))=\{0\}$), but I am struggling to find a proof of this.
I believe any cusp form must be of the form $\lambda(3E_{2,2}-E_{2,4})$ for some $\lambda\in \mathbb{C}$, but I do not manage to obtain nothing from that.
I am searching for a proof that uses the fact that $E_{2,2}$ and $E_{2, 4}$ generate $M_2(\Gamma_0(4))$.
Quite analogously to the $SL_2(\mathbb{Z})$ case, where the argument principle gives the relation $\nu_i/2+\nu_\omega/3+\sum \nu_z+\nu_\infty=\mathrm{weight}/12$, using the standard explicit fundamental domain, for the conjugate $\Gamma_\theta$ of $\Gamma_o(4)$, there is $\nu_i/2+\nu_\infty+\sum \nu_z = \mathrm{weight}/4$. Thus, analogous to the corresponding discussion for $SL_2(\mathbb Z)$, for suitably low weights there can be no vanishing at infinity.
For example, my notes https://www-users.cse.umn.edu/~garrett/m/mfms/notes_2013-14/10_thetas_equi.pdf treat this (well-known) situation.
Yes, this is an application of Riemann-Roch in low genus, so gives explicit formulas easily.