I don't have an answer, but I can give some context for the positions of zeros (which you may already know). The dimension formula for the space of modular forms of weight k is essentially (# of zeros) + 1 - restrictions. Here is the formula for even k, taken from Shimura (Introduction to the Arithmetic Theory of Automorphic Functions); I believe he has formulas for odd weight as well.

The space of modular forms of even weight *k* on a subgroup *G* of SL_{2}(ℤ) with $I=[SL_2(\mathbb{Z}): \pm G]$, r_{2} order-2 elliptic points, and r_{3} order-3 elliptic points has dimension
$d = Ik/12 + 1 -(k/4 -\lfloor k/4 \rfloor )r_2 - (k/3 - \lfloor k/3 \rfloor)r_3 - g$.

The r_{2} and r_{3} terms come from the zeros required by the valence formula. The -*g* can be seen as coming from conditions imposed by Abel's Theorem. What's left gives us the degrees of freedom in placing zeros and getting a unique modular form, up to a constant multiple.

In your example, specifying a cusp form of a certain weight leaves a 1-dimensional space, and we know where all the zeros are. For weight 12 forms on SL_{2}(ℤ), we have genus 0 and don't need zeros at the elliptic points, so we have a 2-dimensional space. Specifying a zero at any point gives us a unique form: a zero at the cusp gives $\Delta$, a zero (of degree 3) at the order-3 point gives $E_4^3$, a zero (of degree 2) at the order-2 point gives $E_g^2$. A zero anywhere else comes from some linear combination, but I don't know how to relate the linear combination to the position of the zero.

It's more complicated in the higher genus case, because in a space with dimension *d=n+1-g*, you can place *n* zeros anywhere you want, but there are *g* additional zeros in the fundamental domain, somehow determined by the locations of the first *n*.