I'm trying to get a more intuitive understanding for the following theorem.
Theorem. $ $ Every continuous odd map $f∶S^2⟶\mathbb{R}^2$ has a zero in $S^2.$
Am I right in thinking that this theorem implies that $f$ deforms $S^2$ (through $\mathbb{R}^3$) onto the closure of a simply connected subset of $\mathbb{R}^2$ containing the origin $(0, 0)$? So, basically, $f$ is a projection of $S^2$ and its image is always homeomorphic to $D^2$ (the closed unit disc centred at the origin)?
No, you are not right. For instance, projecting the unit sphere in $\Bbb R^3$ orthogonally to the $z$-axis is an odd map, and there are plenty of odd maps from the interval $[-1,1]$ on the $z$-axis to the plane. Composing gives you a collection of odd maps from the sphere to the plane. Very few of these give an image homeomorphic to $D^2$, and many of them don't have simply connected images either.