Let $E/K$ be an extension of algebraic number fields and let $\mathcal{O}_E$ and $\mathcal{O}_K$ be the respective rings of integers.
For any prime ideal $\mathfrak{P} \subset \mathcal{O}_E$, $\mathfrak{p} := \mathfrak{P} \cap \mathcal{O}_K$ is a prime ideal in $\mathcal{O}_K$, and $\mathfrak{P} \mid \mathfrak{p}$.
We call $\mathcal{O}_K/\mathfrak{p}$ the residue field of $K$ with respect to the prime $\mathfrak{p} \subset \mathcal{O}_K$.
Similarly, $\mathcal{O}_E/\mathfrak{P}$ is the residue field of $E$ with respect to the prime $\mathfrak{P} \subset \mathcal{O}_E$.
Then if $\mathfrak{P} \mid \mathfrak{p}$, we define the corresponding residue field extension as:
$$(\mathcal{O}_E/\mathfrak{P})/(\mathcal{O}_K/\mathfrak{p}).$$
But how do we make sense of this?
The elements of $\mathcal{O}_E/\mathfrak{P}$ are of the form $x\ (\textrm{mod}\ \mathfrak{P})$ where $x \in \mathcal{O}_E$, whereas the elements of $\mathcal{O}_K/\mathfrak{p}$ are of the form $x\ (\textrm{mod}\ \mathfrak{p})$ where $x \in \mathcal{O}_K$.
So in what sense can $\mathcal{O}_E/\mathfrak{P}$ be said to be an extension of $\mathcal{O}_K/\mathfrak{p}$, or equivalently: How can $\mathcal{O}_K/\mathfrak{p}$ be said to be a subfield (or even a subset) of $\mathcal{O}_E/\mathfrak{P}$?
Addendum:
Could one say that every $x \in \mathcal{O}_K$ that is a representative of an element in $\mathcal{O}_K/\mathfrak{p}$ is also a representative of an element in $\mathcal{O}_E/\mathfrak{P}$?
Best Answer
Let $f \colon A \to B$ be any morphism of commutative rings, and let $J$ be any ideal of $B$. Set $I = f^{-1}(J)$, and let $\pi_{I} \colon A \to A/I, \pi_{J} \colon B \to B/J$ denote the canonical quotient morphisms. Then the composition $$A \xrightarrow{f} B \xrightarrow{\pi_{J}} B/J$$ has precisely kernel $I$, and so factors through a (unique) injection $\overline{f} \colon A/I \to B/J$ such that $\overline{f} \circ \pi_{I} = \pi_{J} \circ f$ by the universal property of the quotient.
Specializing to your situation, $A = \mathcal{O}_{K}, B = \mathcal{O}_{E}, J = \mathfrak{P}, I = \mathfrak{p}$, and $f$ is just the inclusion $\mathcal{O}_{K} \hookrightarrow \mathcal{O}_{E}$.