Let $A$ be a commutative ring with unity. In Ravi Vakil's notes he defines $O_{Spec A} (D(f))$ to be the localization of $A$ at the multiplicative set of all functions that do not vanish outside of $V(f)$. Note $D(f)$ is all the primes that do not contain $f \in A$ and $V(f)$ is all the primes that contain $f$.
He then gives an exercise that states: show that the natural map $A_f \rightarrow O_{Spec A} (D(f))$ is an isomorphism. What does it mean by "natural" here? Does it mean more than just the most obvious map kind of thing? Thank you!
In theory there could be lots of maps $A_f \to \mathcal{O}_{\operatorname{Spec A}}(D_f)$. Of course, there's an obvious one, and that's what he means by "the natural map."
"Natural" means something like "depending on a uniform procedure and no arbitrary choices." It has a technical meaning in category theory, though generally cooking up the relevant categories and functors is left to the reader who wants to do it.
I'll illustrate in this case. Since we are concerned with a ring $A$ and an element $f \in A$, we should probably consider the category $\mathcal{C}$ with $$\operatorname{Ob}(\mathcal{C}) = \{(A, f) \; | \; A \text{ is a ring and } f \in A \}$$ $$\operatorname{Hom}_{\mathcal{C}}\bigg((A, f), (B, g)\bigg) = \{ \varphi : A \to B \; | \; \varphi(f) = g \text{ and $\varphi$ is a ring homomorphism}\}$$
Then we have functors $$F, G : \mathcal{C} \to \mathsf{Ring}$$ where $$F(A, f) = A_f \qquad \text{and} \qquad G(A, f) = \mathcal{O}_{\operatorname{Spec A}}(D_f).$$ (Of course, I'm cheating a bit here, because a functor maps both objects and morphisms, and I haven't actually specified what these functors do to morphisms. Hopefully that's obvious from context.)
A natural transformation $$\eta : F \to G$$ is, by definition, a set of morphisms $$\eta_{A, f} : F(A, f) \to G(A, f),$$ one for each $(A, f) \in \operatorname{Ob}(\mathcal{C})$, such that the diagram
$$\begin{array}{c} (A,f) & & F(A, f) = A_f & \xrightarrow{\eta_{A,f}} & G(A, f) = \mathcal{O}_{\operatorname{Spec} A}(D_f) \\ \downarrow_{\varphi} & & \downarrow_{F(\varphi)} & & \downarrow_{G(\varphi)} \\ (B, g) & & F(B, g) = B_g & \xrightarrow{\eta_{B,g}} &G(B, g) = \mathcal{O}_{\operatorname{Spec} B}(D_g) \end{array}$$ always commutes. (Sorry for the awful diagram; I don't know the best way of doing them here.)
Saying that $\eta$ is a natural transformation between $F$ (the ring localization) and $G$ (taking sections over an elementary open set) means, essentially, that $\eta$ "plays well" with the ways that we convert a ring homomorphism $A \to B$ into a map $A_f \to B_g$ and into a map of sections.
Caveat: This is meant to somehow uniquely identify the map $A_f \to \mathcal{O}_{\operatorname{Spec} A}(D_f)$ under discussion, but I don't think it's really the case that there's only one such natural map.
(I mean, obviously there isn't, because the zero map is also natural, but we can rule that out by throwing "nonzero" somewhere.)[Thanks Georges Elencwajg]. I haven't considered this situation in particular, and I'm not an expert on this sort of thing, but it's often the case that given one natural map you can build some others out of it. So, really, the best thing to do here is probably to replace "natural" with "obvious."