(1) I think saying $x\mod\mathcal{P}(L)$ isn't quite right, we should probably say $x\mod L$. Since $L$ is an additive subgroup of $\mathbb{R}^n$, we can take the quotient group $\mathbb{R}^n/L$, and this quotient is basically $\mathcal{P}(L)$ (with some extra structure). So you're right that this has uncountably many points; however, $x\mod \mathcal{P}(L)$ (or really, $x\mod L$) is just a single point in $\mathcal{P}(L)$, so we don't need a superposition over the full set $\mathcal{P}(L)$.
Instead, given some fixed $y\in\mathcal{P}(L)$, we want a superposition over all $x\in \mathbb{Z}^n$ such that $x\equiv y\mod L$. This set is countable (in fact, it's equal to $L+y$), which is still infinite but, as Regev argues, we can get close enough to this superposition since everything is weighted by the discrete Gaussian $\rho_{\sqrt{2}r}$.
(2) Consider that we can take any vector $x$ and compute two vectors $v\in L$ and $y\in \mathcal{P}(L)$ such that $v+y=x$. This is an easy task if we do not care about the length of $y$; one way is to just find a set of coefficients to express $x$ as a linear combination of some basis of $L$, then just round the coefficients to the nearest integer. That linear combination gives $v$, the difference gives $y$. No matter how we do this we know that $x\equiv y\mod L$, since $v\in L$.
This can be done out-of-place, so suppose that we let $x\mod L$ be the vector in $\mathcal{P}(L)$ that we obtain from such a procedure called on $x$. Then we map
$$ \sum_{x\in\mathbb{Z}^n}\rho_{\sqrt{2}r}(x)\vert x\rangle\mapsto \sum_{x\in\mathbb{Z}^n}\rho_{\sqrt{2}r}(x)\vert x\rangle\vert x\mod L\rangle$$
If we measure the second register, we get some value $y\in\mathcal{P}(L)$, and we collapse the state onto only those $x$ such that $x\equiv y\mod L$. However, as argued above, that's precisely equal to the set $L+y$. Hence, the remaining state is
$$ \sum_{x\in\mathbb{Z}^n:x\equiv y\mod L}\rho_{\sqrt{2}r}(x)\vert x\rangle = \sum_{x\in L+y}\rho_{\sqrt{2}r}(x)\vert x\rangle$$
Notice that there are only certain values in $\mathcal{P}(L)$ that we could measure, since we will only measure $y$ such that $y\equiv x\mod L$ for some $x$ in our original superposition. Since $x\in\mathbb{Z}^n$ for all states in the original superposition, we know that $y+L\subseteq \mathbb{Z}^n$, since those are the only values of $y$ we could possibly measure.
