This is only a partial answer. Regarding the last part of your question:

Every (connected) finite branched cover $f: V\rightarrow \mathbb{D}$ of the disc is isomorphic to one of the form $p_n: \mathbb{D}\rightarrow \mathbb{D}$, where $p_n(z) = z^n$ and $n\in \mathbb{Z}^+$. In particular, $f^{-1}(0)$ consists of a single point since $z^n$ has a unique zero.

To get a better idea of why only the map $z\rightarrow z^n$ appears, note that every such branched covering arises from an unbranched covering $g:X\rightarrow \mathbb{D}^*$ which one can also show is of the form $p_k: \mathbb{D}^* \rightarrow \mathbb{D}^* $ up to isomorphism. The original branched covering $V\rightarrow \mathbb{D}$ is obtained by completing, i.e. "adding an extra point" to $\mathbb{D}^*$ and extending $g$ so that it takes the value $0$ on this new point. So $f^{-1}(0)$ consists of a single point (of "multiplicity" $k$) by construction. Now, if you don't require $V$ to be connected, you'll end up with possibly several disjoint copies of this situation, namely $N$ discs $\mathbb{D}_j$ equipped with maps $z\rightarrow z^{n_j}$. Then $f^{-1}(0)$ consists of $N$ points $v_j$ with multiplicities $n_j$. This is the situation you'll find when analyzing the preimage of a small embedded disc under a non-constant map of Riemann surfaces.

In higher dimensions, branched coverings are more difficult to understand. One useful fact is that such maps are proper (by definition). So if $V,W$ are smooth, then the Proper Mapping Theorem applies. In particular, $f(V\backslash V_0)$ is an analytic hypersurface in $W$ (since the ramification divisor $V\backslash V_0$ is an analytic hypersurface in $V$ - see Griffiths & Harris).

Edit: One nice reference for the basics of branched covers is the book "Holomorphic Functions to Complex Manifolds" by Grauert and Fritzsche.