In [OEIS A124796](https://oeis.org/A124796) I considered a similar problem of computing the coefficients of $(\partial_z\circ M_g)^n$, where $M_g$ is the operator of multiplying by $g(z)$.

It turns out that the coefficients represent generalized Stirling numbers indexed by infinite vectors of nonnegative integers ${\cal S}([k_0,k_1,k_2,\dots])$, where ${\cal S}([k_0,k_1,0,0,\dots]) = S(k_0+k_1+1,k_0+1)$ are conventional Stirling number of the 2nd kind. 

The expansion for $(\partial_z\circ M_g)^n$ is given by
$$(\partial_z\circ M_g)^n = \sum_{k_0+k_1+\dots=n\atop k_1+2k_2+\dots\leq n} {\cal S}([k_0,k_1,\dots]) \prod_{i\geq 0} (\partial_z^i g(z))^{k_i}\cdot \partial_z^{n-(k_1+2k_2+\dots)}.$$

The coefficients satisfy a recurrence relation:
$${\cal S}([k_0,k_1,\dots]) = {\cal S}([k_0-1,k1,\dots]) + (k_0+1){\cal S}([k_0-1,k_1,\dots]) + \sum_{i\geq 2} (k_{i-1}+1) {\cal S}([k_0-1,k_1,...,k_{i-2},k_{i-1}+1,k_i-1,k_{i+1},\dots])$$
with ${\cal S}([0,0,\dots])=1$ and ${\cal S}([k_0,k_1,\dots])=0$ as soon as any $k_i<0$.

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Going back to the original question, it concerns $(M_g\circ\partial_z)^n = M_g\circ (\partial_z\circ M_g)^{n-1}\circ \partial_z$ and thus
$$(M_g\circ\partial_z)^n = g(z)\cdot \sum_{k_0+k_1+\dots=n-1\atop k_1+2k_2+\dots\leq n-1} {\cal S}([k_0,k_1,\dots]) \prod_{i\geq 0} (\partial_z^i g(z))^{k_i}\cdot \partial_z^{n-(k_1+2k_2+\dots)}.$$