I have the following shape: $A := $ {$(z,w) \in \mathbb{C}^2 \mid \text{Arg}(z) + \text{Arg}(w) = 0 \text{ mod } 2\pi$} inside $\mathbb{C}^2$.
I know that $A \cap S^3 \subseteq S^3 \hookrightarrow \mathbb{C}^2$ is supposed to be an annulus (where $S^3$ is the set of pairs $(z,w) \in \mathbb{C}^2$ such that $|z|^2 + |w|^2 = 1$). (here for an actual picture of it, embedded in $S^3$)
It might even be that $A \subseteq \mathbb{C}^2$ already looks like an annulus by itself, without even having to look at it inside $S^3$, but I'm not sure about that bit. So I'm just mentioning it.
The farthest that I can get is that $A =$ {$(z,w) \in \mathbb{C}^2 \mid z \underset{\,\,\,\,\,\,\mathbb{R}}{\propto} \bar{w}$} (that is, before intersecting it with $S^3$), but this doesn't help me visualizing it much. And even less, see why it should be an annulus?
The question of why this is an annulus is relatively easy to answer on a mathematical level. Perhaps knowing why will aid you in visualizing this annulus, but frankly visualization is kind of a subjective process and I'm not sure I can say much more than answering your question why. Nonetheless I'll say a few words about visualization at the end.
To answer the why, the first point to make is that for all $(z,w) \in A \cap S^3$, the inequality $|w|^2 + |z|^2 = 1$ implies that $|w| \le 1$ and that $|z| \le 1$. The second point is that, because of the requirement that both $\text{arg}(w)$ and $\text{arg}(z)$ be defined, neither $w=0$ nor $z=0$ is allowed; from this it follows that neither $|w|=1$ nor $|z|=1$ is allowed.
Taking these points altogether it follows that, as $(w,z) \in A \cap S^3$ varies, each of $w$ and $z$ take values in the following open annulus in $\mathbb C$: $$\mathcal A = \{r e^{i\theta} \mid 0 < r < 1 \} $$ Define a projection $p : A \cap S^3 \to \mathcal A$, where $p(z,w)=z$. This projection is continuous (obviously). Also $p$ is one-to-one and onto: for each $z = r e^{i\theta} \in \mathcal A$ there exists a unique $w \in \mathcal A$ such that $|z|^2 + |w|^2 = 1$ and $\arg(z) + \arg(w) = 0$, namely $$w = w(re^{i\theta}) = \sqrt{1-r^2} \, e^{-i\theta} $$ Finally, this formula shows that $w(z)$ is a continuous function of $z$, and therefore the inverse function $\mathcal A \mapsto A \cap S^2$ defined by $z \mapsto (z,w(z))$ is continuous. This proves that $p : A \cap S^3 \to \mathcal A$ is a homeomorphism.
As for visualization, it depends on how you like to visualize the unit sphere $S^3$. One popular method is to use stereographic projection which maps $S^3 - \{(1+0i, 0+1i)\}$ diffeomorphically to $\mathbb R^3$. One could certainly use sterographic projection to plot the graph $\{(z,w(z))\}$ of the function $w=w(z)=w(re^{i\theta})$ discussed above, as a subset of $\mathbb R^3$. It seems quite likely to me that this is the intention of the hand-drawn picture that you linked in your post.