We work over $\mathbb{C}$. Let $X$ be a normal projective variety, let $E$ be a rank $2$ vector bundle on $X$ and denote by $\pi: \mathbb{P}\to X$ the associated $\mathbb{P}^1$-bundle. I know that it holds that $\text{Pic}(\mathbb{P}(E))=\pi^*(\text{Pic}(X))\oplus \mathbb{Z} \mathcal{O}_{\mathbb{P}(E)}(1)$.
I know this is a very stupid question, but does a similar result also hold at the level of the group of Weil (or Cartier) divisors? For istance, if I consider a prime divisor $H$ on $\mathbb{P}(E)$, can I always write it as a sum $H=\pi^*D_X+\mathcal{O}_{\mathbb{P}(E)}(k)$, where by $D_X$ I denote a Weil divisor on $X$ (and therefore $\pi^*(D_X)$ is a codimension 1 subvariety of $\mathbb{P}(E)$), and $k\in \mathbb{Z}$?
I apologize in advance, I knwo I have a lot of confusion right now and I would like to understand.
Let $X^0 \subset X$ be the open complement of the singular locus. Then $$ \mathrm{Cl}(X) = \mathrm{Cl}(X^0) = \mathrm{Pic}(X^0) $$ (the first holds because $\mathrm{codim}(X \setminus X^0) \ge 2$ and the second holds because $X^0$ is smooth). Similarly, $$ \mathrm{Cl}(\mathbb{P}_X(E)) = \mathrm{Cl}(\mathbb{P}_{X^0}(E\vert_{X^0})) = \mathrm{Pic}(\mathbb{P}_{X^0}(E\vert_{X^0})). $$ Now the Picard formula that you know implies the same formula for the class group.