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: $${\cal S}([k_0,k_1,\dots]) = {\cal S}([k_0-1,k_1,\dots]) + (k_0+1){\cal S}([k_0,k_1-1,k_2,\dots]) + \sum_{i\geq 1} (k_i+1) {\cal S}([k_0-1,k_1,...,k_{i-1},k_i+1,k_{i+1}-1,k_{i+2},\dots])$$ 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.
UPDATED. The original question concerns $(M_g\circ\partial_z)^n = M_g\circ (\partial_z\circ M_g)^{n-1}\circ \partial_z$. Hence, \begin{split} (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)} \\ &= \sum_{k_0+k_1+\dots=n\atop k_1+2k_2+\dots\leq n} {\cal C}([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)}, \end{split} where ${\cal C}([k_0,k_1,\dots]) = {\cal S}([k_0-1,k_1,\dots])$ for $k_0\geq 1$, and ${\cal C}([0,k_1,k_2,\dots])=0$ except for ${\cal C}([0,0,0,\dots])=1$. In fact, the formula with coefficients ${\cal C}([k_0,k_1,\dots])$ holds even for $n=0$.
Correspondingly, we have a recurrence relation: $${\cal C}([k_0,k_1,\dots]) = {\cal C}([k_0-1,k_1,\dots]) + k_0{\cal C}([k_0,k_1-1,k_2,\dots]) + \sum_{i\geq 1} (k_i+1) {\cal C}([k_0-1,k_1,...,k_{i-1},k_i+1,k_{i+1}-1,k_{i+2},\dots]).$$ Then the generating function $$F(z_0,z_1,\dots) := \sum_{k_0,k_1,\dots\geq 0} {\cal C}([k_0,k_1,\dots]) \prod_{i\geq 0}z_i^{k_i}$$ satisfies the differential equation: $$F = 1 + z_0 F + z_0 \sum_{i\geq 0} z_{i+1}\partial_{z_i} F.$$ If $F_n$ is the restriction of $F$ to the terms of degree $n$, then $F_0=1$ and for $n>0$: $$F_n = z_0 F_{n-1} + z_0 \sum_{i\geq 0} z_{i+1}\partial_{z_i} F_{n-1}.$$$$F_n = z_0 F_{n-1} + z_0 \sum_{i=0}^{n-2} z_{i+1}\partial_{z_i} F_{n-1}.$$
Examples.
- $F_1 = z_0$
- $F_2 = z_0^2 + z_0z_1$
- $F_3 = z_0^3 + 3z_0^2z_1 + z_0z_1^2 + z_0^2z_2$
- $F_4 = z_0^4 + 6 z_0^3 z_1 + 7z_0^2z_1^2 + z_0z_1^3 + 4z_0^3z_2 + 4z_0^2z_1z_2 + z_0^3z_3$
As expected, the coefficients in $F_n(z_0,z_1,0,0,\dots)$ are Stirling numbers of the 2nd kind.
It's worth to notice that for $g(z)=z$, we have $(M_g\circ\partial_z)^n = \sum_{k=0}^n S(n,k) z^k \partial_z^k$, which is essentially an umbral Touchard polynomial.