A question on a projective bundle $\mathbb{P}(L\oplus \mathcal{O}_X)$ Let $X$ be a complex manifold and $L$ a line bundle on it. Define $Y:=\mathbb{P}(L\oplus \mathcal{O}_X)$ be the projective bundle over $X$. Here is a statement I don't understand:

The summands $L$ and $\mathcal{O}_X$ respectively determine divisors on $Y$, each of which is isomorphic to $X$.

How can one define such divisors on $Y$? 
 A: This is a rather standard fact. 
Let $\mathscr{E}$ be a rank $2$ locally free sheaf on $X$ and let $\pi \colon \mathbb{P}(\mathscr{E}) \to X$ be the corresponding $\mathbb{P}^1$-bundle.  Then there is a natural correspondence between sections $\sigma \colon X \to \mathbb{P}(\mathscr{E})$ of $\pi$ and suriections $\mathscr{E} \to \mathscr{L} \to 0$, where $\mathscr{L}$ is a line bundle on $X$. Such a correspondence is given by $\mathscr{L}= \sigma ^* \mathscr{O}_{\mathbb{P}(\mathscr{E})}(1)$.
So in your setting the two quotients $L$ and $\mathscr{O}_X$ of $\mathscr{E}=L \oplus \mathscr{O}_X$ determine two sections of $\pi \colon \mathbb{P}(\mathscr{E}) \to X$. Of course, any section is a divisor in $\mathbb{P}(\mathscr{E})$ which is isomorphic to $X$ via the projection $\pi$.
A: For any locally free sheaf $\mathcal{E}$ on $X$, the projective bundle $\mathbb{P}(\mathcal{E}\oplus \mathcal{O}_X)$ possesses an open subset $\mathbb{V}(\mathcal{E})$ whose closed complement is isomorphic to $\mathbb{P}(\mathcal{E})$ and corresponds to the divisor associated to the sheaf $\mathcal{O}_X(1)$.
In your case $\mathrm{dim}(\mathbb{P}(\mathcal{L}\oplus \mathcal{O}_X))= 1$, so $\mathbb{P}(\mathcal{L})$ is the divisor at infinity (a point) and $\mathbb{V}(\mathcal{L})$ is the complementary line, see Fulton-Lang "Riemann-Roch Algebra" IV, 1. Moreover, if you consider the canonical map $\pi \colon \mathbb{P}(\mathcal{L}\oplus \mathcal{O}_X)) \to X$ then $\pi^*\mathcal{L}$ will correspond to a divisor of degree one in your projective line, again isomorphic to the divisor associated to the sheaf $\mathcal{O}_X(1)$.
