Sections of $\mathcal{O}_{\mathbb{P}^1}(n)$ are homogeneous degree $n$ polynomials in $x_0,x_1$. You can consider the two sections $x_0^n,x_1^n,$ and construct the morphism $$g:\mathbb{P}^1\rightarrow\mathbb{P}^1\times\mathbb{P}^2,\:[x_0,x_1]\mapsto ([x_0,x_1][0,x_0^n,x_1^n])$$.
You have also the morphism
$$e:\mathbb{P}^1\rightarrow\mathbb{P}^1\times\mathbb{P}^2,\:[x_0,x_1]\mapsto ([x_0,x_1][1,0,0])$$
Note that the images of these two morphism are disjoint. The image of $e$ is going to be the unique curve with self-intersection $-n$. Let $[y_0,y_1,y_2]$ be the homogeneous coordinates on $\mathbb{P}^2$. The fiber of $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n))$ over the point $[x_0,x_1]\in\mathbb{P}^{1}$ is exactly the line
$$F_{x_0,x_1} = \left\langle [x_0^n,x_1^n,0],[0,0,1]\right\rangle = \{x_1^ny_1-x_0^ny_2 = 0\}.$$
Then $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n))$ is the surface $\mathbb{F}_n$ in $\mathbb{P}^{1}\times\mathbb{P}^2$ defined by $x_1^ny_1-x_0^ny_2 = 0$.

Another method: let $E= e(\mathbb{P}^1)$ and $F$ a fiber. Then $E+F$ represents the sections of type
$$p:\mathbb{P}^1\rightarrow\mathbb{P}^1\times\mathbb{P}^2,\:[x_0,x_1]\mapsto ([x_0,x_1][p_n(x_0,x_1),x_0^n,x_1^n])$$
where $p_n(x_0,x_1)$ is a homogeneous polynomial of degree $n$. The curve $E$ is given by $y_1 = y_2 = 0$ in $\mathbb{F}_n$, therefore $E^{2} = -n$. The linear system $|E+nF|$ has degree zero on $E$, one on $F$ and $n$ on $E+F$. This means that it maps $\mathcal{F}_n$ to a surface $C$ in $\mathbb{P}^{n+1}$ whith a unique singular point (corresponding to the contraction of $E$) and whose hyperplane section is a rational normal curve of degree $n$. Then $C$ is a cone over a rational normal curve of degree $n$ and the vertex is a singularity of type $\frac{1}{n}(1,1)$. The surface $\mathbb{F}_n$ is the blow up of the vertex of the cone which is isomorphic to the scroll $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n))$.

Finally, in a less direct way, one could observe that both $\mathbb{F}_n$ and $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n))$ are geometrically ruled surfaces over $\mathbb{P}^1$, and that both contain a unique curve with negative self-intersection $-n$. This is enough to conclude that $\mathbb{F}_n\cong \mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n))$.

`$x_0$`

and`$x_1$`

by 7, for example. – Serge Lvovski Feb 26 '13 at 7:08