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 $3\times 3$ matrice 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.

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.