This answer addresses the second question: "If we assume the fibers of $\pi$ are curves of genus $0$, can we embed $X$ into $\mathbb{P}^2_S$?"

The answer to this is also no (providing there is a singular fibre).

Let $\pi: X \to \mathbb{P}^1$ be a conic bundle surface over an algebraically closed field $k$, i.e. $X$ is regular and every fibre is isomorphic to a plane conic (possibly singular).

Consider the relative anticanonical bundle $\omega_{X/\mathbb{P}^1}^{-1}$. This is very ample when restricted to each fibre, and the pushforward to $\mathbb{P}^1$ is a vector bundle $V$ of rank $3$. We obtain an embedding

$$X \to \mathbb{P}(V)$$ which respects $\pi$, where $\mathbb{P}(V)$ denotes the corresponding $\mathbb{P}^2$-projective bundle over $\mathbb{P}^1$.

Now every vector bundle on $\mathbb{P}^1$ splits as a direct sum of line bundles, so we can write

$$V = \mathcal{O}(a_1) \oplus \mathcal{O}(a_2) \oplus \mathcal{O}(a_3).$$

We obtain the trivial projective bundle if and only if $a_1 = a_2 = a_3$. So we just need to give an example where this doesn't hold.

Firstly, if every fibre is smooth, then $X$ is a ruled surface. By the classification of ruled surfaces $X$ is a Hirzebruch surface $\mathbb{F}_n$ for some $n$. But $\mathbb{F}_n$ embeds into $\mathbb{P}^2 \times \mathbb{P}^1$ as $$x_0^n y_0 = x_1^n y_1.$$ So to get a counter-example we need to consider a conic bundle with a singular fibre.

I take $X$ to be a smooth cubic surface in $\mathbb{P}^3$. For any line $L \subset X$, there is an associated conic bundle given by taking the residual intersection of the pencil of planes through $L$. This has exactly 5 singular fibres as $L$ meets exactly 10 other lines of $X$.

Now take $Y$ a conic bundle surface in $\mathbb{P}^2 \times \mathbb{P}^1$. This has bidegree $(2,d)$ for some $d$. Considering the discriminant of the associated quadratic form one sees that this has $3d \neq 5$ singular fibres, as required.

Incidently, one can show that any smooth cubic surfaces embeds into $\mathbb{P}(V)$ where $V = \mathcal{O}(0) \oplus \mathcal{O}(0) \oplus \mathcal{O}(1).$

This answer addresses the second question: "If we assume the fibers of $\pi$ are curves of genus $0$, can we embed $X$ into $\mathbb{P}^2_S$?"

The answer to this is also no (providing there is a singular fibre).

Let $\pi: X \to \mathbb{P}^1$ be a conic bundle surface over an algebraically closed field $k$, i.e. $X$ is regular and every fibre is isomorphic to a plane conic (possibly singular).

Consider the relative anticanonical bundle $\omega_{X/\mathbb{P}^1}^{-1}$. This is very ample when restricted to each fibre, and the pushforward to $\mathbb{P}^1$ is a vector bundle $V$ of rank $3$. We obtain an embedding

$$X \to \mathbb{P}(V)$$ which respects $\pi$, where $\mathbb{P}(V)$ denotes the corresponding $\mathbb{P}^2$-projective bundle over $\mathbb{P}^1$.

Now every vector bundle on $\mathbb{P}^1$ splits as a direct sum of line bundles, so we can write

$$V = \mathcal{O}(a_1) \oplus \mathcal{O}(a_2) \oplus \mathcal{O}(a_3).$$

We obtain the trivial projective bundle if and only if $a_1 = a_2 = a_3$. So we just need to give an example where this doesn't hold.

Firstly, if every fibre is smooth, then $X$ is a ruled surface. By the classification of ruled surfaces $X$ is a Hirzebruch surface $\mathbb{F}_n$ for some $n$. But $\mathbb{F}_n$ embeds into $\mathbb{P}^2 \times \mathbb{P}^1$ as $$x_0^n y_0 = x_1^n y_1.$$ So to get a counter-example we need to consider a conic bundle with a singular fibre.

I take $X$ to be a smooth cubic surface in $\mathbb{P}^3$. For any line $L \subset X$, there is an associated conic bundle given by taking the residual intersection of the pencil of planes through $L$. This has exactly 5 singular fibres as $L$ meets exactly 10 other lines of $X$.

Now take $Y$ a conic bundle surface in $\mathbb{P}^2 \times \mathbb{P}^1$. This has bidegree $(2,d)$ for some $d$. Considering the discriminant of the associated quadratic form one sees that this has $3d \neq 5$ singular fibres, as required.

Incidently, one can show that any smooth cubic surface embeds into $\mathbb{P}(V)$ where $V = \mathcal{O}(0) \oplus \mathcal{O}(0) \oplus \mathcal{O}(1).$