If I have an extension $L/K$ of number fields, then I can take the inclusion $\mathcal{O}_K \hookrightarrow \mathcal{O}_L$ and get a morphism of "curves" $\operatorname{Spec} \mathcal{O}_L \to \operatorname{Spec} \mathcal{O}_K$.
Can I do something like this for an arbitrary field extension? Or at least for a tower $F \subseteq K \subseteq L$ with $\operatorname{trdeg}_F K = \operatorname{trdeg}_F L$?
a) No, I don't think you can do that: if $F=K=\mathbb R\subset L=\mathbb C$, what on earth would be $\mathcal O_\mathbb R$ ?
b) Here is a comment on this rhetorical question.
Strangely, $\mathbb R$ is the fraction field of some non-trivial subring $A\subsetneq \mathbb R$:
Take a transcendence basis $(r_i)_{i\in I}$ of $\mathbb R$ over $\mathbb Q$, then consider the ring $P=\mathbb Q[r_i|i\in I]$ and take for $A$ the integral closure of $P$ in $\mathbb R$.
Then $A$ is not a field because $P$ isn't (Atiyah-Macdonald, Proposition 5.7) but $\operatorname {Frac(A)}=\mathbb R$ (Atiyah-Macdonald, Proposition 5.12) .
c) So, why not take $\mathcal O_\mathbb R=A ?$
Because $A$ is incredibly not canonical and I don't even think that such an $A$ could be proved to exist without the axiom of choice.