$\def\Pic{\text{Pic}}$You are writing things in a very sophisticated way. I'm going to respond to the question at the top but not try to follow all of the thoughts in the middle. For a scheme $X$, I'll write $\Pic(X)$ for the abelian group of isomorphisms classes of line bundles on $X$.
(1) The functor $U \mapsto \Pic(U)$ is not a sheaf.
The "locality" condition in the definition of a sheaf says that, if $X = U \cup V$ is an open cover, and $A$ and $B$ are line bundles on $X$ with $A|_U \cong B|_U$ and $A|_V \cong B|_V$, then $A \cong B$. This is the opposite of how line bundles work!
Take $X = \mathbb{P}^1$ and let $U$ and $V$ be the standard cover of $\mathbb{P}^1$ by two $\mathbb{A}^1$'s. Every line bundle becomes trivial on $U$ and on $V$, but $\mathbb{P}^1$ has nontrivial line bundles.
If you for want $X$ to be affine as well (since your original question talked about $\text{Aff}$) take $X = \text{Spec} \mathbb{Z}[\sqrt{-5}]$, $U = \text{Spec} \mathbb{Z}[\sqrt{-5}, 2^{-1}]$ and $V = \text{Spec} \mathbb{Z}[\sqrt{-5}, 3^{-1}]$. Again, $\mathbb{Z}[\sqrt{-5}]$ is not a UFD but $\mathbb{Z}[\sqrt{-5}, 2^{-1}]$ and $\mathbb{Z}[\sqrt{-5}, 3^{-1}]$ are, so there are nontrivial line bundles on $X$ which become trivial on each of $U$ and $V$.
(2) The stronger statement that the stalks of $U \mapsto \Pic(U)$ are all trivial. I find the language of stalks on the big Zariski site awkward, so let's just talk about open sets on a fixed scheme $X$. Then $U \mapsto \Pic(U)$ is a presheaf on $X$.
Let $p$ be any point of $X$. Then the stalk of this presheaf is $\lim_{\to} \Pic(V)$ where the limit is over all open neighborhoods $V$ of $p$. Let $U$ be any open neighborhood of $p$, and let $L$ be any line bundle on $U$. Then there is some neighborhood $V$ of $p$ such that $L|_V$ is trivial (that's what a line bundle means!). So $[L]$ maps to $1$ in $\Pic(V)$. So every class becomes trivial in the stalk.
Well, if you are restricting your attention to schemes, there is simply not any better option. One way of saying this is that $V(xy)$ is the smallest closed subscheme that contains both $V(x)$ and $V(y)$. This just amounts to the algebraic fact that any element of $\mathbb{C}[x,y]$ which is divisible by both $x$ and $y$ must be divisible by $xy$. Another way to make this precise is to say that $V(xy)$ is the pushout of $V(x)$ and $V(y)$ over their intersection $V(x,y)$ in the category of schemes. So, if you restrict your attention to the the world of schemes, there just does not exist any better candidate to call the union of the two lines.
Your observation that this is not true in the larger sheaf topos is closely related to the fact that the algebraic fact above is not stable under base-change: if you map $\mathbb{C}[x,y]$ to some other $\mathbb{C}$-algebra, then the images of $x$ and $y$ may no longer have the property that any element divisible by both $x$ and $y$ is divisible by $xy$. In the context of schemes, this is the fact that the operation of taking unions of closed subschemes is not stable under base-change. Explicitly, for instance, if you pull back to the diagonal line $V(x-y)$ in $\mathbb{A}^2$, then $V(x)$ and $V(y)$ both become just the reduced point at the origin, but $V(xy)$ becomes a non-reduced thickened version.
Best Answer
Consider $T=\Bbb P^1\times S$. Then $X_T=X\times_S S\times \Bbb P^1=\Bbb P^1\times X$, and denoting $p: \Bbb P^1\times X\to \Bbb P^1$ the projection, we have that $p^*\mathcal{O}(1)$ is a nontrivial invertible sheaf on $X_T$.