This may not be the answer you're looking for but it is how I would approach this theorem. Convergence matters rely on Cauchy's theorem anyway so why don't use that directly in a proof?
Let $c < r < R < 1$ and define $f_r: \{z \mid r < |z| < 1\} \times \mathbb{P}^{n-1} \to \mathbb{C}$ by
$$
f_r(z, w) = \frac{1}{2 \pi i}\int_{|v|=r} \frac{f(v, w)}{v - z}dv.
$$
Then $f_r$ is holomorphic and vanishes if $|w_k| > c$ for some index $k$ since $v \mapsto f(v, w)$ is then holomorphic within the contour $|v|=r$. This implies that $f_r$ is identically zero. In particular $f$ has the following representation on $\{z \mid r < |z| < R\} \times \mathbb{P}^{n-1}$
$$
f(z, w) = \frac{1}{2 \pi i} \int_{|v| = R} \frac{f(v, w)}{v - z}dv.
$$
But the right hand side defines a holomorphic function on $\{z \mid |z| < R\} \times \mathbb{P}^{n-1}$.
Actually, it's not so hard to find a solution of $\frac{\partial}{\partial\overline{z}}G =\delta$. You have the right thing there already, modulo normalisation. We have, in the sense of distributions, for any $\varphi\in \mathscr{D}(\mathbb{C})$,
$$\begin{align}
\left(\frac{\partial}{\partial\overline{z}}\frac{1}{z-w}\right)[\varphi]
&= - \int_{\mathbb{C}} \frac{\partial\varphi}{\partial\overline{z}}(z)\frac{1}{z-w}\,d\lambda\\
&= -\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert z-w\rvert < R} \frac{\partial\varphi/\partial\overline{z}(z)}{z-w}\,d\lambda
\end{align}$$
where $\lambda$ is the Lebesgue measure on $\mathbb{C}$ and $R$ is sufficiently large, so that the support of $\varphi$ is contained in $D_R(w)$. Now write the integral with differential forms, that is, replace $d\lambda$ with $dx\wedge dy$, and write the latter as $\frac{1}{2i}d\overline{z}\wedge dz$. We get
$$\begin{align}
\left(\frac{\partial}{\partial\overline{z}}\frac{1}{z-w}\right)[\varphi] &= -\lim_{\varepsilon\to 0} \frac{1}{2i}\int_{\varepsilon < \lvert z-w\rvert < R} \frac{\partial\varphi/\partial\overline{z}(z)}{z-w}\,d\overline{z}\wedge dz\\
&= -\frac{1}{2i}\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert z-w\rvert < R} d\left(\frac{\varphi(z)}{z-w}\, dz\right) \tag{Stokes}\\
&= \frac{1}{2i}\lim_{\varepsilon\to 0} \int_{\lvert z-w\rvert = \varepsilon} \frac{\varphi(z)}{z-w}\,dz\\
&= \frac{1}{2i}\lim_{\varepsilon\to 0} \left(\int_{\lvert z-w\rvert = \varepsilon} \frac{\varphi(w)}{z-w}\,dz + \int_{\lvert z-w\rvert =\varepsilon} \frac{\varphi(z)-\varphi(w)}{z-w}\,dz \right)\\
&= \pi \varphi(w),
\end{align}$$
since by the differentiability of $\varphi$ in $w$, the integrand of the second integral in the penultimate line remains bounded as $\varepsilon\to 0$, and so the second integral is $\leqslant C\cdot 2\pi\varepsilon$ in absolute value by the ML-inequality. That means
$$\frac{1}{\pi}\frac{\partial}{\partial\overline{z}} \frac{1}{z-w} = \delta_w.$$
Best Answer
$\def\dbar{\overline{\partial}}\def\zbar{\overline{z}}$This isn't geometric, but here is a non-rigorous argument suggesting this formula that I gave when I taught this lemma.
Consider the case where $g$ is the $\delta$ function $\delta_w$ at $w$. Can we find a smooth function $f$ such that, in some sense, $\tfrac{\partial f}{\partial \zbar} = \delta_w$? In other words, for any disc $D$, we want $$\int_D \frac{\partial f}{\partial \zbar} d(\mathrm{Area}) = \begin{cases} 1 & w \in D \\ 0 & w \not\in D \end{cases}.$$ The wedge product $d z \wedge d \zbar$ is $2i$ times the area form, so we want $$\int_D \frac{\partial f}{\partial \zbar} dz \wedge d \zbar = \begin{cases} 2 i & w \in D \\ 0 & w \not\in D \end{cases}.$$ Since $df = \tfrac{\partial f}{\partial z} dz + \tfrac{\partial f}{\partial \zbar} d\zbar$, we can rewrite the left hand side as $$- \int_D df \wedge dz = - \int_D d \left(f dz \right) =- \int_{\partial D} f dz.$$ So we want a smooth function $f$ such that $$-\int_{\partial D} f dz = \begin{cases} 1 & w \in D \\ 0 & w \not\in D \end{cases}.$$ The obvious choice is $f(z) = \tfrac{1}{\pi (w-z)}$.
Now, consider a general function $g(z)$. We have $$g(z) = \int_w g(w) \delta_w(z) \ d(\mathrm{Area}).$$ Since $\dbar$ is linear and we "have" $\tfrac{\partial}{\partial \zbar} \tfrac{1}{\pi(w-z)} = \delta_w(z)$, we should have $$f(z) = \frac{\partial}{\partial \zbar} \int_w g(w) \frac{1}{\pi(w-z)} \ d(\mathrm{Area}).$$
The last formula is correct for any compactly supported smooth $g$ and is the formula you ask for, up to the relation $dw \wedge d \overline{w} = (2 i) d (\mathrm{Area})$.
PS: You might wonder what happens if we replace $\tfrac{1}{\pi (w-z)}$ by some other meromorphic $h(w,z)$ with a pole of residue $\tfrac{-1}{\pi}$ at $w$ and no other poles. If $h$ blows up to higher order at $w$, then $\int_w g(w) h(w,z) d(\mathrm{Area})$ won't converge absolutely even for compactly supported $g$, so we will have to specify what the integral means more carefully, which is a pain. So we want $h(w,z)$ to be $\tfrac{1}{\pi (w-z)}$ plus an entire function of $z$. If that entire function is nonzero, the result is still right for any compactly supported $g$. Making the simplest choice $\tfrac{1}{\pi (w-z)}$ rather than, say $\tfrac{1}{\pi (w-z)} + e^{wz}$, has the advantage of making the integral still converge if $|g(z)| = O(1/|z|^{1+\epsilon})$. I don't think that is a big advantage though; we still eventually want to prove the Dolbeault lemma for all $g$ and, as far as I know, that still needs a partition of unity argument, which I did in my next lecture.