Classification of $3$-pointed rational curves I tried to prove that $\mathbb P^1 \setminus \{0,1,\infty\}$ is the fine moduli space for the moduli problem, which assigns to a scheme $S$ the set of (isomorphim classes of) $4$-pointed rational curves over $S$. By these I mean smooth, proper morphisms
$$
f: X \rightarrow S
$$
whose geometric fibers are (actually smooth, projective) rational curves, together with four pairwise disjoint sections $\sigma_1,...,\sigma_4$.
For this, I would like to prove the following statement, whose proof (or reference with proof, or at least hints) is what I am kindly asking for

If $f: X \rightarrow S$ is a $3$-pointed rational curve, then there is a unique isomorphism of $f$ with the trivial $\mathbb P^1$-bundle
  $$
S \times \mathbb P^1 \rightarrow S
$$
  such that the three sections $\sigma_1,\sigma_2,\sigma_3$ correspond to the constant sections $0,1,\infty$.

This seems to be well-known, but the only reference I could find are the notes, p.16, where the above statement is not proved. The part about the three sections corresponding to $0,1,\infty$ is probably straightforward, once one knows the triviality of the bundle.
PS: The above notes also contain the following, quite interesting classification results for $1$- resp. $2$-pointed curves, which I state rather roughly. If anyone of you knows something about them, I would really appreciate it.

$1$-pointed rational curves (over $S$) correspond to projective bundles $\mathbb P(\mathcal E)$, where $\mathcal E$ is a rank $2$ vector bundle on $S$, while $2$-pointed rational curves correspond to bundles as above, which are split.

 A: Let $f: X \rightarrow S$ be a flat family of smooth, rational, projective curves. I will assume the following fact: if $f$ has at least one section, then $X \cong \mathbb{P}(\mathcal{E})$, where $\mathcal{E}$ is a rank $2$ vector bundle on $S$ (follow the idea in the proof of the proposition V.2.2 in Hartshorne's Algebraic Geometry, which consider the case where $S$ is a surface).
Assume that $f$ has at least two sections, which are disjoint. This sections correspond to surjections $\mathcal{E} \twoheadrightarrow L_{1}$ and $\mathcal{E} \twoheadrightarrow L_{2}$, where $L_{1}$ and $L_{1}$ are line bundles on $S$. Assuming that $S$ is integral, the kernels $K_{1}$ and $K_{2}$ of these morphisms are both line bundles. Since the two sections are disjoint, you have an injection $K_{1} \oplus K_{2} \hookrightarrow \mathcal{E}$. Since $K_{1} \oplus K_{2}$ and $\mathcal{E}$ are both of rank $2$, we have the surjectivity, and therefore $\mathcal{E} \cong K_{1} \oplus K_{2}$.
Now assume that $f$ has at least three sections, which are disjoint. This guarantee that we do not have two of them mapping a point $s \in S$ to the zero element of the vector space $\mathcal{E}_{s}$. In other words, we have a frame for $\mathcal{E}$, and therefore, $\mathcal{E}$ is trivial, that is, $\mathbb{P}(\mathcal{E}) \cong S \times \mathbb{P}^{1}$. Basically, what I am using here is the fact that a vector bundle of rank $n$ is trivial if and only if has $n$ global sections linearly independent.
I think you have noted that in Kock-Vainsencher's An Invitation to Quantum Cohomology - Kontsevich's Formula for Rational Plane Curves, the universal family for $4$-pointed rational curves is presented in Example 1.1.4.
