Let $f:X\to Y$ be morphism of schemes, and $\mathscr{L}$ be inventible sheaf over $Y$.
Now we consider $\mathscr{L}$ as line bundle $L\to Y$ as Hartshorne II.Ex.5.18.
On the other hand, we get inventible sheaf $f^{*} \mathscr{L}$ over $X$, and as above we think $f^{*} \mathscr{L}$ as line bundle $M\to X$.
Now does fiber product $X\times_Y L$ correspond to $M $ as scheme？
If $\DeclareMathOperator{\Spec}{Spec} f: X:=\Spec(B) \rightarrow S:=\Spec(A)$ and if $L\in Pic(A)$ it follows $\DeclareMathOperator{\Sym}{Sym}\mathbb{V}(L^*):=\Spec(\Sym_A^*(L^*))$, hence there are isomorphisms
$$\mathbb{V}((f^*L)^*):=\Spec(\Sym_B^*((B\otimes_A L)^*) \cong \Spec(B\otimes_A \Sym_A^*(L^*)) \cong X\times_S \mathbb{V}(L^*).$$
Note For a locally trivial $A$-module $L\DeclareMathOperator{\Hom}{Hom}$ it follows $\Hom_B(B\otimes_A L, B) \cong B\otimes_A \Hom_A(L,A)$, hence
$$(B\otimes_A L)^* \cong B\otimes_A L^*.$$
Moreover
$$ \Sym_B^*((B\otimes_A L)^*) \cong B\otimes_A \Sym_A^*(L^*).$$