In Vakil's "the rising sea" book, the base gluability axiom is defined as:
"(Let $F$ be a sheaf on base {$B_i$}) If $B = \cup B_i$, and we have $f_i \in F(B_i)$ such that $f_i$ agrees with $f_j$ on any basic open set contained in $B_i \cap B_j$ (i.e.,$res_{B_i,B_k}f_i = res_{B_j,B_k}f_j$ for all $B_k \subset B_i \cap B_j$) then there exists $f \in F(B)$ such that $res_{B,B_i}f = f_i$ for all i."
Why the base gluability axiom require the condition "for any basic open set contained in $B_i \cap B_j$" but not "it exists one basic open set contained in $B_i \cap B_j$"?
What if we replace "any" by "one"?
Notice that we only use the "one" condition to prove the extension problem of base sheaf, rather the "any" condition.
When we replace for "any" by "one", then there will be situations where the data $(B_i,f_i)$ satisfies the condition even though you can not glue them.
E.g. consider the topological space $\mathbb R$ which its standard topology. That topology has a basis consisting of open intervals. Let $F$ be the presheaf of functions on $\mathbb R$. This means that for each open subset $U$ of $\mathbb R$ the set $F(U)$ is the set of all set-theoretic functions $U\to \mathbb R$. Consider $B_1 = (-\infty,1)$ and $B_2=(-1,+\infty)$. Then $B = B_1\cup B_2 = \mathbb R$. Let $f_1$ be the function $f_1\in F(B_1)$ which is zero everywhere, except at the origin where $f_1(0) = 1$. Let $f_2\in F(B_2)$ be the function which is zero everywhere, except at the origin, where $f_2(0) = -1$.
Now $f_1$ and $f_2$ agree on some basic open subset contained in $B_1\cap B_2 = (-1,1)$, for example on $(0,1)$, but not on all of them. You also see that $f_1$ and $f_2$ can not possibly be glued to a function $f$ on $B_1\cup B_2$. So with your definition the presheaf $F$ would not be a sheaf.