Let $\pi:\mathbb{P}^1\times\mathbb{P}^1\rightarrow\mathbb{P}^2$ be the double covering branched along a conic $C\subset\mathbb{P}^2$ such that $\pi^{-1}(C)$ is the diagonal.
Is $\pi_*\mathcal{O}_{\mathbb{P}^1\times\mathbb{P}^1}(1,0)\simeq\mathcal{O}_{\mathbb{P}^2}^{\oplus 2}$?
(I suspect this because the dimensions of the global sections of both sides match)
Denote $E=\pi_*\mathcal{O}_{\mathbb{P}^1\times \mathbb{P}^1}(1,0)$ for simplicity. First we show that $E$ admits a nowhere vanishing global section.
We know that for a point $p\in \mathbb{P}^2$ the fiber of $E$ is given by $E\otimes \kappa(p) = H^0(\pi^{-1}(p),\mathcal{O})$, and since the global sections of $E$ are the global sections of $\mathcal{O}(1,0)$ we just need to find a line $\ell$ of this ruling such that for every $p\in \mathbb{P}^2$ the line does not contain $\pi^{-1}(p)$.
We can do this geometrically: look at the map $\pi$ as the projection from a point $q\in \mathbb{P}^3$ to an hyperplane $H$, restricted to a smooth quadric $Q\subseteq \mathbb{P}^3$ such that $q\notin Q$. In this setting $C=H\cap Q$. Then we take any line $\ell$ of the ruling: for any point $p\in \mathbb{P}^2$, the fiber consists of the intersection between the line $\ell(p,q)$ and $Q$. These are two points, counted with multiplicity, so that if $\ell$ contains these points, it would intersect $\ell(p,q)$ in two points, meaning that $\ell(p,q)=\ell$, but this is absurd since $q\notin \ell$.
This proves that $E$ admits a nowhere vanishing global section $\sigma\in H^0(E)$, which corresponds to an exact sequence $$ 0 \to \mathcal{O}_{\mathbb{P}^2} \overset{\sigma}{\to} E \to C \to 0 $$ The fact that $\sigma$ never vanishes means that $C$ is a line bundle, and moreover taking global sections we get the exact sequence $$ 0 \to H^0(\mathcal{O}_{\mathbb{P}^2}) \to H^0(E) \to H^0(C) \to 0 $$ since $H^1(\mathcal{O}_{\mathbb{P}^2})=0$. In particular, we get $h^0(C)=h^0(E)-h^0(\mathcal{O}_{\mathbb{P}^2})=2-1=1$. This implies $C\cong \mathcal{O}_{\mathbb{P}^2}$, since this is the unique line bundle on $\mathbb{P}^2$ with a one-dimensional space of global sections.
To conclude, we just observe that $$ \text{Ext}^1(\mathcal{O}_{\mathbb{P}^2},\mathcal{O}_{\mathbb{P}^2}) = H^1(\mathcal{O}_{\mathbb{P}^2})=0 $$ so that every extension of $\mathcal{O}_{\mathbb{P}^2}$ by $\mathcal{O}_{\mathbb{P}^2}$ is trivial, and in particular $E\cong \mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2} $.