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 SL2(ℤ) with $I=[SL_2(mathbb{Z}): pm G]$, r2 order-2 elliptic points, and r3 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 r2 and r3 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 SL2(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.
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