In OEIS 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 ${\cal S}([k_0,k_1,k_2,\dots])$ with a finite number of nonzero components, 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: $${\cal S}([k_0,k_1,\dots]) = {\cal S}([k_0-1,k_1,\dots]) + (k_0+1){\cal S}([k_0,k_1-1,\dots]) + \sum_{i\geq 2} (k_{i-1}+1) {\cal S}([k_0-1,k_1,...,k_{i-2},k_{i-1}+1,k_i-1,k_{i+1},\dots])$$ with ${\cal S}([0,0,\dots])=1$, and ${\cal S}([k_0,k_1,\dots])=0$ when any $k_i<0$ or when $k_1+2k_2+\dots>k_0+k_1+k_2+\dots$ (in other words, $k_2+2k_3+\dots > k_0$). In particular, the sum in the r.h.s. of the recurrence relation consists of just a finite number of nonzero terms.
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 $$(M_g\circ\partial_z)^n = g(z)\cdot \sum_{k_0+k_1+\dots=n-1\atop k_1+2k_2+\dots\leq n-1} {\cal S}([k_0,k_1,\dots]) \prod_{i\geq 0} (\partial_z^i g(z))^{k_i}\cdot \partial_z^{n-(k_1+2k_2+\dots)}.$$
It's worth to notice that for $g(z)=z$, we have $(M_g\circ\partial_z)^n = \sum_{k=0}^n S(n,k) x^k \partial_z^k$, which is essentially an umbral Touchard polynomial.