I think I have a rough solution to my question now. We may write $f$ as
$$ f=h\cdot g=h\cdot (w-b_1(z))...(w-b_d(z)) $$
where the $b_i(z)$ are the "roots" of $g$. They depend on $z$ since the coefficients of $g$ depend on $z$. It follows that the zero locus of $f$ contains the hypersurfaces $w=b_i(z)$ for all $i=1,...,d$.
What suggests now the idea of projecting into the plane ($w=0$) is that we can regard each hypersurface as a function of $w$ in coordinates $(z,w)$. So they look like surface in this coordinates.
A way to think geometrically in the case $z\in\mathbb{C}$ goes as follows. If I draw the real parts of each function in the plane $(Re(z),Re(w))$ what we will see is lines representing a function, and this lines intersect only where $g$ has multiple roots. This lines form a branched covering of the horizontal axis which is $w=0$.
The discriminant function of a polynomial $g$ is precisely a holomorphic function whose zero locus is the set of multiple roots. Therefore the conclusion that the zero locus forms a branch cover when projecting to $(w=0)$ is now evident, the branch locus of this projection is the zero locus of the determinant of $g$.
I now think the conclusion of the statement is trivial, but what I needed to "see" is that the roots can be seen as functions $w=b_i(z)$ ,that $g$ has multiple roots exactly where its discriminant vanishes (I did not know this) and that this function is analytic.
You forgot to assume that $f$ is not identically zero. Here is one way to argue:
Step 1. I will start with some general topology observations. Let $X$ be a topological space and $A\subset X$ a subset. One says that $A$ does not locally separate $X$ if for every $x\in X$ there is a neighbourhood $W_x$ of $x$ in $X$ such that $W_x\setminus A$ is connected.
I will now assume, mostly for convenience, that $X$ is metrizable. (In the case of interest, $X$ will be an open subset of ${\mathbb C}^n$.)
Lemma. Suppose that $X$ is connected, $A\subset X$ is closed, $int(A)=\emptyset$ and $A$ does not locally separate $X$. Then $X\setminus A$ is connected.
Proof. Let $V$ be a connected component of $X\setminus A$. I claim that $cl_X(V)=X$. If not, then, by connectedness of $X$, there is a boundary point $x$ of $cl_X(V)$ in $X$, and, hence, a sequence
$x_i\in X\setminus cl(V)$ converging to $x$. Since $A$ is closed and has empty interior in $X$, the sequence $x_i$ can be chosen so that $x_i\notin A$ for each $i$. Let $W_x$ denote a neighborhood of $x$ such that
$W_x\setminus A$ is connected. Since $x$ is in the closure of $V$, the intersection $W_x\cap V\subset W_x\setminus A$ is also nonempty. Furthermore, for some $i$, $x_i$ is in $W_x\setminus A$. By the connectedness of $W_x\setminus A$, it follows that $x_i$ lies in the connected component $V$ of $X\setminus A$. A contradiction.
Thus, $cl_X(V)=X$. Since $cl_X(V)\subset V \cup A$ (as $V$ is a connected component of $X\setminus A$), it follows that $X=V\cup A$, i.e. $X\setminus A$ is connected. qed
I will now turn to your question.
Let $X$ be an open connected subset of ${\mathbb C}^n$, $f: X\to {\mathbb C}$ a nonconstant holomorphic function, $A:= f^{-1}(0)$. I will be using the fact that $A$ has empty interior in $X$ (already proven).
Step 2. Now, prove that $A$ does not locally separate $X$. Namely, given $x\in X$, take $W_x$ to be an open ball centered at $x$ and contained in $X$. Verify connectedness of $W_x\setminus A$ by looking at the intersections of complex lines $L$ with $W_x$: For each pair of distinct points $p, q$ in $W_x\setminus A$ take the complex line $L$ containing $p, q$. Then $f|_{L\cap X}$ is nonconstant, i.e. $L\cap A$ is a discrete subset of $L\cap X$. (I will leave it to you as an exercise.) From this, conclude path-connectedness of $L\cap (W_x\setminus A)$ and, hence, path-connectedness of $W_x\setminus A$. (Again, an exercise.)
- Now, combine Lemma and and Step 2 to conclude that the zero-level set of a nonconstant holomorphic function does not separate the domain, as required.
Best Answer
I have found a solution to the problem by reading the proof of the WPT (as in Proposition 1.1.6 of Huybrechts' Complex Geometry).
We first note that the sections $f(z_1,0) \equiv 0$ and $f(0,z_2)= z_2^2$ so we can only single out the coordinate $z_2$ as "the good coordinate". In order to construct the Weierstrass polynomial, the proof of theorem considers
$$ p_{z_1}(z_2) = \prod_{i=1}^{d} (z_2 - a_i(z_1)), $$ where $a_i(z_1)$ are the roots of $f_{z_1}(z_2)$ that restrict to the original root $z_2=0$ when the parameter $z_1=0$. The proof of the theorem gives an argument that shows that this is a Weierstrass polynomial in which the coefficients $\alpha_j$ are holomorphic, and written in terms of sums of the form $$ \sum_{i=1}^{k}a_i(z_2)^k. $$ which then proves that are holomorphic by using the residue theorem.
In my particular case I can calculate the zeros of $f_{z_1}(z_2)$ by the quadratic formula. Factoring out $z_2$: $$ f_{z_1}(z_2) = z_2(z_1^3 + z_1 + (z_1^2+z_1)z_2 + z_1 z_2^2) $$ and by the quadratic formula I get \begin{align*} a_0(z_1) & = 0,\\ a_1(z_1) &= \frac{-1-z_1^2 + \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2 z_1},\\ a_2(z_1) & = \frac{-1-z_1^2 - \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2 z_1}. \end{align*} It is easy to see that $a_2(z_1)$ is not well defined for $z_1\rightarrow 0$, and that $a_1(z_1)\rightarrow 0$ for $z_1\rightarrow 0$. We consider then the Weierstrass polynomial $$ p_{z_1}(z_2) = z_2 \left( z_2 - \frac{-1-z_1^2 + \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2 z_1}\right). $$ This implies that the holomorphic function $h(z_1,z_2)$ is precisely \begin{align*} h(z_1,z_2) & = f(z_1,z_2)/p_{z_1}(z_2) = \frac{z_1 z_2 (z_2-a_1(z_1))(z_2-a_2(z_1))}{z_2(z_2-a_1(z_2))} \\ & = z_1 (z_2 -a_2(z_1)) = z_1z_2 + \frac{1+z_1^2 + \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2} \end{align*} which, note, does not vanish at the origin $h(0,0)= 1$ and is thus a unit in the ring of germs $\mathcal{O}_{\mathbb{C}^{n},0}$.