The question pertains to the notes found at: https://ocw.mit.edu/courses/mathematics/18-101-analysis-ii-fall-2005/lecture-notes/lecture6.pdf.
In the proof, $U_1$ is defined to be a $\delta$ open rectangle and $V$ a $\delta/4$ open ball. The inverse theorem states that $U_1$, $V$ are both neighbourhoods of $a$ and $f(a)$ respectively. When applying the inverse theorem, does this mean that $U_1$ has to be a rectangle and $V$ an open ball?
The theorem says nothing about how the neighborhoods must be constructed. It only says that if the theorem is applicable then there exists a diffeomorphism between $U_1$ and $V$ (which implies that both are open sets). In many cases it's not trivial to invert the function explicitly (think of a very complicated function $f$) and specify the corresponding neigborhoods $U_1$ and $V$ - you only know that they do exist.
Let's consider the following example: $$f:\mathbb{R}^2\to \mathbb{R}^2, f{x \choose y} = {2x^2 \choose 3y^2} .$$ Can we apply the inverse function theorem at ${0 \choose 0}$? No, but at point ${1 \choose 1}$ we can. We can even calculate the inverse function explicitly: $$f^{-1}:\mathbb{R}^2\to \mathbb{R}^2, f^{-1}{x \choose y} = {\frac{1}{\sqrt{2}}\sqrt{x} \choose \frac{1}{\sqrt{3}}\sqrt{y}} .$$
How about the sets $U_1$ and $V$? Well, regarding $U_1$ we can take any open set as long as $x>0$ and $y>0$. We could either define a rectangle as in the proof or some other weird set as long as it is open and $x>0$ and $y>0$. Note that we don't specify $V$ directly but $V$ is obtained implicitly by defining $U_1$. $V$ is always open due to continuity of $f$.