From Vakil's Foundations of Algebraic Geometry:
The open sets of the distinguished affine base are the affine open subsets of $X$. We have already observed that this forms a base. But forget that fact. We like distinguished open sets $\operatorname{Spec} A_f \hookrightarrow \operatorname{Spec} A$, and we don’t really understand open embeddings of one random affine open subset in another. So we just remember the "nice" inclusions.
13.3.1.Definition. The distinguished affine base of a scheme $X$ is the data of the affine open sets and the distinguished inclusions.
Vakil writes that this a "not a base in the usual sense."
Is this not a base in the usual sense?
If $X$ is a topological space, a collection of open subsets $\mathcal B$ forms a base if every open subset of $X$ is a union of elements of $\mathcal B$.
Let $U$ be an open subset of a scheme $X$. Let $p \in U$. Then $p$ is in some affine open subset of $X$, say $\operatorname{Spec} A$. Then $p \in U \cap \operatorname{Spec} A$, which is open in $\operatorname{Spec} A$, hence $p \in \operatorname{Spec} A_f$ for some $f \in A$. $\operatorname{Spec} A_f$ is an open subset of an open subset, hence open in $X$. So, $p \in \operatorname{Spec} A_f \subset U$.
Best Answer
As he says in the text, Vakil is describing a (non-full) subcategory of the category of open sets of the scheme $X$. Namely, he is considering the category whose objects are open affine subsets $U$ and where there is a single morphism $U \to V$ if and only if $U$ is a distinguished open affine of $V$
The reason this is not a base in the usual topological sense is because it could be that we have two affine opens $U \subseteq V$, with a distinguished affine open $W$ in $U$ that is not distinguished open affine in $V$. In other words, in the distinguished affine base we “forget” that $W$ is a subset of $V$. The upshot is that to construct a quasicoherent sheaf on a scheme, we don’t need to remember arbitrary inclusions of open sets.