Surface of type $(2,2)$ on the Segre cubic scroll $\mathbb{P}^1 \times \mathbb{P}^2 \subset \mathbb{P}^5$

Here is another approach. Take a smooth surface $X$ of type $(d,2)$ in $S$. The general fiber of the projection $\pi:X\rightarrow \mathbb{P}^1$ is a smooth conic. So $X$ is rationally connected and since $X$ has dimension $2$ it is rational. This does not depend on $d$.

The bi-homogeneous polynomial cutting out $X$ in $S$ of the the following form:

$$a_{0,0}(u,v)x^2+a_{0,1}(u,v)xy+a_{0,2}(u,v)xz+a_{1,1}(u,v)y^2+a_{1,2}(u,v)yz+a_{2,2}(u,v)z^2$$

where $[u:v]$ and $[x:y:z]$ are homogeneous coordinates on $\mathbb{P}^1$ and $\mathbb{P}^2$ respectively, and the $a_{i,j}(u,v)$ are homogeneous polynomials of degree $d$. The matrix of the conic $C_{u,v}$ over the point $[u:v]\in\mathbb{P}^1$ is a $3\times 3$ matrix whose entries are homogeneous polynomials of degree $d$. So its determinant is a homogeneous polynomial of degree $3d$. The singular conics in the conic bundle correspond to the zeros of the determinat. So you have $3d$ singular conics.

This surface $X$ satisfies $p_g(X)=q(X)=0$. Moreover, if $H_X$ is the hyperplane section, by adjunction we find $$K_X^2=2, \quad K_X H_X=-4.$$

Therefore no multiple of the canonical divisor can be effective, in particular $P_2(X)=q(X)=0$ and $X$ is rational by Castelnuovo criterion.

In fact, $X$ is a conic bundle with six singular fibres, that turns out to be the blow-up at six points of the quadric $\mathbb{F}_0=\mathbb{P}^1 \times \mathbb{P}^1$. This is Example 1.9, p. 236 in

S. Di Rocco, K. Ranestad: On surfaces in $\mathbb{P}^{6}$ with no trisecant lines (http://dx.doi.org/10.1007/BF02384319), Ark. Mat. 38, No. 2, 231-261 (2000). ZBL1035.14011.