I forgot put that $D$ was bounded

Yes you are right $\mathbb{C}^{\times}$ is a counter-example, so I'll redit my question

put parenthesis around $\Delta(\tau)/\Delta(d \tau)$.

@Michael, I just thought about it but an example of a rational function $z$ would be something like $\prod_{d|N}\frac{\Delta(\tau)}{\Delta(d\tau)}^{a_d}$ with these 3 conditions (1) $\sum_d a_d=0$, (2) $\sum_d da_d=0$ and $a_d=a_{N/d}$ for all $d|N$.

Right, right so this space will be incredibly huge! So too construct your function $z$ you might try to take the quotient of two linearly independent "rational "Poincare series of the same weight. Something like $\sum_{\gamma\in \Gamma_N} H(\gamma(z))$ for two suitable $H$'s should work.

Thanks Michael for your answer, but I'm wondering if there is some kind of combinatorial-arithmetic description of this vector space of 1-forms. Do you think it will be exhausted by linear combinations of modular forms attached to abelian varieties and Eisenstein series?

@Noam, if $N$ is composite then you could consider something like $\sum a_d d E_2(dz)$ with the two conditions that $\sum a_d d=0$ and $a_{d}=a_{N-d}$. Then I think this gives something in $V$.

yes right $E_2(z)-NE_2(Nz)$ does not vanish at $\frac{1}{0}$ but I think it vanishes at $\frac{0}{1}$, I have to check...

So @Noam do you think you could use Poincare's trick and average out over the group generated by $z\mapsto z+1$ and $z\mapsto \frac{-1}{Nz}$. Of course, one has to be careful about convergence issues.

Yes I just edited. The transformation $z\mapsto \frac{-1}{Nz}$ is not in the modular group unless $N=1$ but I guess you could work in a bigger group which contains the involution of level $N$.