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EDIT 2: Csar's comment comes very close to answering the question for me, and I will try and spell out the ideas in that comment a little better. I wish Csar had left his comment as an answer so I could select it.

So it seems to me that the most basic reason to think about schemes over a field $k$ is this:

I already spelled out above why $k$-valued points of a scheme are important. But a lot of the time, a morphism from $Spec(k)$ to $X$ will point to a generic point of $X$, not a closed point. Different morphisms can all point to the same generic point. For instance the dual of the any injection $\mathbb{Z}[x,y] \rightarrow \mathbb{C}$ (of which there are many), will all "point" to the generic point of $\mathbb{Z}[x,y]$.

On the other hand if we are looking at $\mathbb{Q}[x,y]$, this is a $\mathbb{Q}$ scheme. $Spec(\mathbb{Q})$ is also a $\mathbb{Q}$ scheme via the identity map. A morphism of $\mathbb{Q}$ schemes from $Spec(\mathbb{Q})$ to $\mathbb{Q}[x,y]$ will correspond to a closed $\mathbb{Q}$ -valued point! So this is a nice geometric reason for looking at schemes over a fixed field $k$: $k$ morphisms of $Spec(k)$ to $X$ correspond to $k$-valued closed points of $X$.

It will take some work for me to internalize the vast generalization that S-schemes entail, but I think this is a good start. Does everything I said above make sense?
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EDIT: The answers I have received so far have been good, but I am looking for something more. I will try to flesh out an example that give the same style of answer that I am asking for:

The notion of a k-valued point: Every point $(a,b)$ of the real variety $x^2 + y^2 - 1 = 0$ has a corresponding evaluation homomorphism $\mathbb{R}[x,y] \rightarrow \mathbb{R}$ given by $x \mapsto a$ and $y \mapsto b$. Since $a^2 + b^2 - 1 = 0$, this homomorphis factors uniquely through $\mathbb{R}[x,y]/(x^2 + y^2 -1)$. So the real valued points of the unit circle are in one to one correspondence with the homomorphisms from $\mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $\mathbb{R}$. Similarly, the complex valued points are in one to one correspondence with the homomorphisms from $\mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $\mathbb{C}$. Actually, points of the unit circle valued in any field $k$ are going to be in one to one correspondence with homomorphisms from $\mathbb{Z}[x,y]/(x^2 +y^2 -1)$ to $k$.

Dualizing everything in sight, we are saying that the $k$- valued points of the scheme $Spec(\mathbb{Z}[x,y]/(x^2 +y^2 -1))$ are just given by homomorphisms from the one point scheme $Spec(k)$ into $Spec(\mathbb{Z}[x,y]/(x^2 +y^2 -1))$.

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Intuition about schemes over a fixed scheme

I am taking a first course on Algebraic Geometry, and I am a little confused at the intuition behind looking at schemes over a fixed scheme. Categorically, I have all the motivation in the world for looking at comma categories, but I would like to make sense of this geometrically.

Here is one piece of geometric motivation I do have: A family of deformations of schemes could be thought of as a morphism $X \rightarrow Y$, where the fibers of the morphism are the schemes which are being deformed, and these are indexed by the scheme Y.

This is all well and good, but I am really interested in Schemes over $Spec(k)$ are thought of as doing geometry "over" $k$. I know that their is a nice "schemeification functor" taking varieties over $k$ to schemes over $k$, but this is somewhat unsatisfying. All that I see algebraically is that $k$ injects into the ring of global sections of the structure sheaf of $X$, but this does not seem like much of a geometric condition...

Any help would be appreciated.