Generic question about commutative algebra

If $K = k$ then the second construction reduces to the first one; the Nullstellensatz implies that maximal ideals correspond exactly to $k$-algebra homomorphisms $A \to k$.

In general the second construction is more general (as it has to be to give all prime ideals instead of just maximal ones). The simplest example to keep in mind is $A = k[x], K = k(x)$ and $A \to K$ the usual inclusion, which defines what is called the generic point of the affine line $\mathbb{A}^1$ and corresponds to the prime-and-not-maximal ideal $(0)$. The second construction also does not require $k$ to be algebraically closed, and can be used to recover maximal ideals by taking $K$ to be a finite extension of $k$ (this follows from a more general version of the Nullstellensatz).

For $A$ a finitely generated $k$-algebra, let $X = \operatorname{m-spec} A$, and let $X(k)$ be the set of $k$-algebra homomorphisms from $A$ to $k$ (we call $X(k)$ the set of $k$-rational points of $X$). There is a natural injective map $X(k) \rightarrow X$ given by sending a $k$-algebra homomorphism to its kernel. One way of stating the Nullstellensatz is that for $k$ algebraically closed, this is a bijection.

Now assume that $k$ is perfect but not necessarily algebraically closed, and let $Y = \operatorname{m-spec} A \otimes_k \overline{k}$. The natural map $\mathfrak m \mapsto \mathfrak m \cap A$ can be shown to define a surjection $Y \rightarrow X$.

The geometric points of $A$ as you call them are the same as $k$-algebra homomorphisms from $A$ into $\overline{k}$, and these are the same as $\overline{k}$-algebra homomorphisms from $A \otimes_k \overline{k}$ into $\overline{k}$. In other words, a geometric point of $A$ is just an element of $Y(\overline{k})$. Now we have a diagram

$$\begin{matrix} X(k) & \subset & Y(\overline{k}) \\ \cap & &|| \\X & \leftarrow & Y\end{matrix}$$

where $Y = Y(\overline{k})$ because of the Nullstellensatz. How do we interpret this diagram? The Galois group $\operatorname{Gal}(\overline{k}/k)$ acts on $Y$ because it acts on $\overline{k}$. It can be shown that $Y \rightarrow X$ is actually the quotient map under this action (this is even a topological quotient if $X$ and $Y$ are taken in the Zariski topologies). Therefore, a maximal ideal of $A$ corresponds to an equivalence class of maximal ideals of $A \otimes_k \overline{k}$ (or geometric points of $A$) under the action of the Galois group, and the $k$-rational points of $X$ are exactly the fixed points of this action.