This is page 10, ex 3.7.
Let $P:=R[[X_1, \ldots,X_n]]$, then $\sum a_{(i)} X_1^{i_1} \ldots X_n^{i_n} \mapsto a_{(0)}$ is a canonical surjective ring map, with kernel $\mathfrak{a}:= \langle X_1 , X_2, \ldots, X_n \rangle $. Hence $P/\mathfrak{a} \cong R$. Now let $\mathfrak{m}$ be any ideal of $R$, and $\mathfrak{n}=\mathfrak{m}P+\mathfrak{a}$. Then $$P/\mathfrak{n} \cong R/\mathfrak{m}.$$
The notes said use the correspondence between ideals of quotients. But this is not clear to me. How does this work in detail?
Denote the projection map you describe by $\pi: P \to R$ with kernel $\ker(\pi)=\mathfrak a$. The correspondence you mentioned states that the ideals of $R$ are in bijection with the ideals of $P$ that contain $\mathfrak a$. The correspondence is more precisely given by mapping an ideal $\mathfrak m$ of $R$ to $$ \pi^{-1}(\mathfrak m)=\mathfrak m P + \mathfrak a=\mathfrak n. $$ Now, consider the map $\varrho:R\to S := R/\mathfrak m$ and let $\varsigma:=\varrho\circ\pi$. Then clearly you have $\varsigma(P)=\varrho(R)$, and we also have $\ker(\varrho)=\mathfrak m$ as well as $$ \ker(\varsigma)=\ker(\varrho\circ\pi)=\pi^{-1}(\ker(\varrho))=\pi^{-1}(\mathfrak m) =\mathfrak n, $$ so you know that $\varsigma(P)\cong P/\mathfrak n$.