There's an MIT OCW package from 2005 for a course in several complex variables available here.
Also, there's a "sketch" of a course given at MIT in 2004 available here.
Lastly, of course there's always this site, where many very smart people are more than happy to help with focused questions! :)
The terms $df_j$, $dz_k$ and $d\omega_j$ are usually seen as one-forms. We can expand the one-form $df_j$ in terms of its parameters,
$$
df_j = \frac{\partial f_j}{\partial \omega_l} d\omega_l + \frac{\partial f_j}{\partial z_l} dz_l,
$$
where summation over the index $l$ is implied. The idea of why it is necessary that
$$dz_k=0 \mbox{ for } k=1,\ldots,n \implies d\omega_j=0 \mbox{ for } j=1,\ldots,m \;, $$
for the implicit function theorem: For the $\omega_j$'s to be uniquely defined in terms of the $z_k$'s, close to the point $(\omega_0,z_0)$, then when there is no change in the $z_k$'s, i.e. $d z_k \cdot v =0$ for every $v \in \mathbb R^n$, there must also be no change in the $\omega_j$'s, i.e. $d \omega_j \cdot u =0$ for every $u \in \mathbb R^m$ - basically the $\omega_j$'s are not free to wonder when the $z_k$'s are fixed.
More concretely we can relate this condition to the standard requirement of the implicit function theorem. To do this, we use $dz_k=0$ for $k=1,\ldots,n$ to reduce the equations $df_j =0$ for $j=1,\ldots, m$ to the following
$$
df_j = \frac{\partial f_j}{\partial \omega_l} d\omega_l =0.
$$
If the determinant $\det\{\partial f_j / \partial \omega_k\}_{j,k} \not = 0$, then we can invert the matrix $\partial f_j / \partial \omega_k$, so let's call $\partial \omega_i / \partial f_l$ its inverse, so that
$$ \frac{ \partial \omega_i}{\partial f_j }\frac{ \partial f_j}{\partial \omega_k} = \delta_{i,k} $$
( summation over $j$ is implied) where $\delta_{i,k} = 1$ if $i=k$ and $0$ otherwise.
Finally, we multiply this inverse with the reduced $df_j$ and sum over $j$ to get
$$
0=\frac{\partial \omega_i} { \partial f_j} df_j= \frac{\partial \omega_i} { \partial f_j} \frac{\partial f_j}{\partial \omega_l} d\omega_l =d\omega_l \delta_{i,l} = d\omega_i \; \mbox{, for } \; i= 1, \ldots, m \,.
$$
So we find $d\omega_i =0$ for every $i$. Working backwards you can see that Hormander's claim implies that $\det\{\partial f_j / \partial \omega_k\}_{j,k} \not = 0$, which is an equivalent necessary condition for the implicit function theorem.
I believe a proof using the $dz_k$'s$ \implies d\omega_j$'s leads to a nice direct proof of the implicit function theorem, and can be quite intuitive (see the 3rd paragraph of this answer). Plus it is suited for use on more abstract manifolds, where the one-forms are well defined.
Hope this helps!
Best Answer
The diagram is an attempt to depict in schematic form some aspects of the fully 4D configuration by keeping track only of the absolute values of the two complex coordinates. (This of course suppresses the angular coordinate of $z_1= r_1 e^{i \theta_1}$ and likewise for $z_2=r_2 e^{i \theta_2}$.) (This trick is motivated by the happy fact that all the relevant pieces of the puzzle can be described by equations involving only absolute values.) The schematic diagram is usually called an absolute space picture.
The ordered pair $(z_1, z_2)$ is represented schematically by $(|z_1|, |z_2|)$.
The distinguished boundary is just the Cartesian product of two circles, $S^1\times S^1$ = $\{ |z_1|=1 \} \times \{ |z_2| =1\}$
which in absolute terms is simply depicted by $(1,1)$. Note that is the corner point of the dark square in absolute space.
In contrast, the normal or ordinary boundary is the union of two parts, one of which is $|z_1|<1 $ crossed with $|z_2|=1$. In absolute space this part is depicted by the upper dark line segment. The other part is the vertical dark line segment.