I'm having trouble understanding why they set the domain and range ($U$ and $V$) as such. Because normally, if it's polar coordinates in one-dimension, you would set $0<\theta<\pi/2$ such that when you take the inverse of the function, it will still remain one-to-one.
How would letting $|\theta|<\pi$ in the domain and exclusion of $(x,0):x≤0$ in the range still let the function be a bijection when we restrict it like that?
First note that by definition bijective = injective + surjective. Currently, the map $f$ is defined as follows, $f:\Bbb{R}^2 \to \Bbb{R}^2$, \begin{equation} f(r,\theta) = \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix} \end{equation} So far there is no restriction on the domain and target space of $f$. It's easy to see that $f$ is not even injective, because $f(1,0) = f(1,2 \pi)$, yet $(1,0) \neq (1,2 \pi)$. Hence it can't be bijective. In this case, injectivity is an issue of the domain of the function, because the trigonometric functions $\sin,\cos$ are periodic with period $2 \pi$. So, what those notes are suggesting is that if you restrict the domain of $f$ to \begin{equation} U_1:= \{(r,\theta) \in \Bbb{R}^2: r>0 \quad \text{and} \quad -\pi < \theta < \pi\} \end{equation} then $f$ is injective. Recall that to prove $f$ is injective on $U$, you need to show that for every pair of points $\xi,\eta \in U$, if $f(\xi) = f(\eta)$, then $\xi = \eta$ (try to prove this yourself).
So, now the function with restricted domain $f: U_1 \to \Bbb{R}^2$ is injective. However, it is not surjective, because its range is not the entire $\Bbb{R}^2$. Note that $f[U_1]$ only contains those points in $\Bbb{R}^2$, which do not lie on the non-negative horizontal axis. In other words, the range of $f:U_1 \to \Bbb{R}^2$ is the set $V_1$ defined by \begin{equation} V_1 := f[U_1] = \Bbb{R}^2 \setminus \{(x,0) \in \Bbb{R}^2 : x \leq 0\} \end{equation} So, now if we restrict the codomain from $\Bbb{R}^2$ to $V_1$, it will become surjective, and hence bijective. Thus, what your notes are claiming is that the function $f:U_1 \to V_1$ is bijective (explicitly write out a proof for injectivity and surjectivity separately if it isn't obvious to you).
Lastly, your notes also claim that $U_1$ and $V_1$ are open subsets of $\Bbb{R}^2$, thereby making $f:U_1 \to V_1$ a $C^1$ diffeomorphism.
Also, as your notes remark, this choice of $U_1$, $V_1$ is not unique. There are other possible restrictions you can place on the domain and codomain of $f:\Bbb{R}^2 \to \Bbb{R}^2$, to make it bijective. For instance, I think the most common one is \begin{align} U_2 &:= \{(r,\theta) \in \Bbb{R}^2 : r> 0 \quad \text{and} \quad 0< \theta < 2\pi\} \\ V_2 &:= f[U_2] = \Bbb{R}^2 \setminus \{(x,0) \in \Bbb{R}^2 : x \geq 0\} \end{align} Another choice (which is geometrically unnatural, so no one would use it in practice) is \begin{align} U_3 &:= \{(r,\theta) \in \Bbb{R}^2 : r< 0 \quad \text{and} \quad 2\pi< \theta < 4\pi\} \\ V_3 &:= f[U_3] = \Bbb{R}^2 \setminus \{(x,0) \in \Bbb{R}^2 : x \leq 0\} \end{align}