Suppose $X$ is an integral scheme, and let $\eta \in X$ be its generic point. Then the local ring $\mathcal{O}_{X,\eta}$ is a field, called the function field of $X$ and denoted $K(X)$.
Why is $K(X)$ called a function field? In what sense (if any) are its elements functions? Is there some conceptual connection between regular functions (on a variety) and $K(X)$?
I'm asking because the name is quite suggestive, but it doesn't seem to follow directly from the definition that $K(X)$ is a field of "functions".
Yes, all elements of $K(X)$ are functions, but there is a caveat:
Caveat : the functions are not defined everywhere
The field $K(X)$ is simply the union of all $\mathcal O(U)$ with $U\subset X$ open, where one identifies $f\in \mathcal O(U)$ with $f|V\in \mathcal O(V)$ whenever $U\supset V$.
So I cheated a bit when I said that rational functions are functions: they are equivalence classes of functions.
But if you are an honest person, you can get along without cheating: in every equivalence class of such pairs $(U,f)$ there is one pair with $U$ maximal and by considering only such maximal pairs you may consider that $K(X)\subset \bigcup _{U \:\text {open}}\mathcal O(U)$ .
So, yes, $K(X)$ consists of regular functions, but these functions are not defined everywhere.
Shafarevich has made the interesting observation that this is a feature which distinguishes algebraic geometry from other geometric theories like topology or differential geometry.