Let A &#x2286;<!-- ⊆ --> B be rings, B integral over A ; let <mi mathvari

To answer your first question, look at the commutative diagram
$\begin{array}{ccc}A& \stackrel{}{\to }& B\\ \downarrow & & \downarrow & \\ A_{\mathfrak{p}}& \stackrel{}{\to }& B_{\mathfrak{p}}\end{array}$
Going the right way, n is contracted to p by assumption. Hence the same holds for the left way. But the map $\mathrm{Spec}{A}_{\mathfrak{p}}\to \mathrm{Spec}A$ is well known to be injective and the sole pre-image of $\mathfrak{p}$ is $\mathfrak{m}$. Thus $\mathfrak{n}$ is contracted to m by the map $A_{\mathfrak{p}}\to B_{\mathfrak{p}}$. Of course the same arguments works for ${\mathfrak{n}}^{\prime }$.
For your second question, note that $B$ is an $A$-module, so this is just the usual notation $M_{\mathfrak{p}}$, when $M$ is an $A$-module and p a prime of $A$.