Let $S=\Bbb{C}[x_0,x_1]$
$\Bbb P^1=\text{Proj}(S)$ breaks up as the union of two distinguished opens, which by proposition $5.11\text{b}$ of Hartshorne has isomorphisms: $$\Bbb P^1= D_+(x_0)\cup D_+(x_1)=\text{Spec}(S_{(x_0)})\cup \text{Spec}(S_{(x_1)})$$
Am I correct that $\text{Spec}(S_{(x_i)})$ take the place of $\Bbb A^1$, and hence am safe to write $\Bbb{A}^1_{x_i}$?
1) The prime ideals of $S_{(x_0)}=\Bbb{C}[x_0,x_1]_{x_0}$ are of the form $\{1,x_0,x_0^2,\cdots\}^{-1}\mathfrak{p}$ where $\mathfrak{p}$ was a prime ideal of $S$. But what does $\{1,x_0,x_0^2,\cdots\}^{-1}(x_0-1,x_1-2)$ mean? Should I see the ideal as a submodule, and take a module localization?
2) Is $S_{(x_0)}$ isomorphic to something more familiar?
3) Is my one coordinate on $\Bbb A^1_{x_0}$, $\frac{x_1}{x_0}$ and my one coordinate on $\Bbb A^1_{x_1}$, $\frac{x_0}{x_1}$?
$1)$ The ring $S_{(x_0)}$ is not $\mathbb{C}[x_0,x_1]_{x_0}$, it is only the elements of degree $0$ in that ring (and so generated by $x_1/x_0$). If $R$ is any ring and $S \subseteq R$ a localizing set, then the prime ideals of $S^{-1}R$ are in bijection with those primes in $R$ whose intersection with $S$ is empty. The notation $S^{-1} \mathfrak{p}$ means the set of elements $x/s \in S^{-1}R$ such that $x \in \mathfrak{p}$. Can you find the primes of $S_{(x_0)}$ now? Note they must be generated by homogeneous elements.
$2)$ I'm not sure anything can get more familiar than $\mathbb{C}[x]$, and as noted in the comments, $S_{(x_0)} \cong \mathbb{C}[x]$.
$3)$ Again, answered in the comments.