Let $\eta \to S$ be the generic point of $S$. The function field of $S$ is exactly the residue field of $\eta$. Let $k(S) \subset K$ be a field extension (no need to assume that $K$ is algebraically closed).

The map $\mathrm{Spec}\, K \to S$ factors through the map $\mathrm{Spec}\, k(S) \to S$. The image of $\mathrm{Spec}\, K \to S$ is thus just $\eta$.

But now you apply a base-change to $K$ and $S \otimes K$ need no longer be irreducible. The image of $\mathrm{Spec}\, K \to S \otimes K$ is a point of $S \otimes K$ which lies above $\eta$; such a point is exactly a generic point of an irreducible component of $S \otimes K$. Different choices of embeddings $k(S) \subset K$ will give rise to different irreducible components, and all the irreducible components will arise this way.

Let $\eta \to S$ be the generic point of $S$. The function field of $S$ is exactly the residue field of $\eta$. Let $k(S) \subset K$ be a field extension (no need to assume that $K$ is algebraically closed).

The map $\mathrm{Spec}\, K \to S$ factors as $\mathrm{Spec}\, K \to\mathrm{Spec}\, k(S) \to S$. The image of $\mathrm{Spec}\, K \to S$ is thus just $\eta$.

But now you apply a base-change to $K$ and $S \otimes K$ need no longer be irreducible. The image of $\mathrm{Spec}\, K \to S \otimes K$ is a point of $S \otimes K$ which lies above $\eta$; such a point is exactly a generic point of an irreducible component of $S \otimes K$. Different choices of embeddings $k(S) \subset K$ will give rise to different irreducible components, and all the irreducible components will arise this way.