Let me turn my comment into an answer.
I claim that for fixed $\mu\in\text{Prob}(\mathbb{C})$ and $f\in C_c(\mathbb{C})$, $$ \int_{\mathbb{C}} f(z)d\mu(z)= \lim_{r\to 0} \frac{1}{\pi r^2} \int_{\mathbb{C}} \mu_w([0,r))f(w) d\lambda(w), $$ where $\lambda$ is the Lebesgue measure on $\mathbb{C}$.
To see this, fix $\epsilon>0$, let $R>0$ be such that the 1-nbd of the support of $f$ is in the $R$-disc, let $r=\min\{\delta(\epsilon/(\pi R^2)),1\}$, where $\delta$ is the one given by the uniform continuity of $f$, and observe that $$ \left|\int_{\mathbb{C}} f(z)d\mu(z)- \frac{1}{\pi r^2} \int_{\mathbb{C}} \mu_w([0,r))f(w) d\lambda(w)\right|\leq $$ $$ \int_{|z|\leq R}d\mu(z)\int_{|w-z|\leq r}d\lambda(w) \frac{|f(z)-f(w)|}{\pi r^2} \leq\epsilon.$$
This gives the required inversion formula, hence answers Q3, as well as Q1.
As for Q2, I am not sure what do you expect. One way to answer would be using the inversion formula given above, that is given a collection $(\mu'_z)_z$ you can define $\mu$ using the above formula and then define the collection $(\mu_z)_z$ as you did. The initial collection is "consistent" iff you got back where you've started.