Degenerations of $\mathbb{P}^1$
Solution 1:
Let $F=X_t$ be a closed fiber of $X\to S$. Let $F_r=F_{\mathrm{red}}$.
Let $k$ be the ground field, supposed to be algebraically closed. The crucial point is to show that $H^0(F, O_F)=k$. Once we prove this, as in the notes you are reading, we have $H^1(F, O_F)=0$ by the invariance of the arithmetic genus in a flat family; the surjective homomorphism $O_F\to O_{F_{r}}$ implies that $H^1(F, O_F)\to H^1(F_{r}, O_{F_r})$ is surjective because $\dim F=1$, hence $F_r$ has arithmetical genus $0$. Then it is easy to show that $F_r$ is a transversal (not necessarily semi-stable) union of projective lines.
The property $H^0(F, O_F)=k$ is called cohomological flatness in dimension $0$. This property is studied in Raynaud's fondamental paper "Spécialisation du foncteur de Picard", IHES 1970. The definition is given in 1.4. Then in 6.1.4 he defines the condition (N) (I don't know why 'N') which is satisfied here by $F=X_t$. In 6.4.2, he shows that $X\to S$ is cohomologically flat if $k$ has characteristic $0$. This proves what you want.
Some remarks:
If $X$ is smooth over $k=\mathbb C$, there is a direct proof of cohomological flatness in Xiao's Lectures notes (somewhere in the begining) and it is probably well known. Unfortunately, one can not reduce the case $X$ normal to the case $X$ smooth just by desingularization.
If $X$ is regular and if the generic fiber of $X\to S$ is also a $\mathbb P^1$ (it is a smooth conic in general if $S$ is just is a regular one-dimensional scheme), then $X$ is obtained by successive blowing ups of closed points starting from $\mathbb P^1_S$, so all irreducible components of $F$ are projective lines.
In general, it is not true (even when $X$ is smooth over $k$ of positive characteristic and $F$ is irreducible) that $H^0(F, O_F)=k$. So in the note you are reading, I think it misses some arguments (it is said that $F$ has no embedded point, so $H^0(F, O_F)⁼k$), unless something in characteristic $0$ is implicit there.
Solution 2:
From an advanced perspective, this follows from a variant of semistable reduction which Harris and Morrison call Nodal Reduction. I quote from their textbook Moduli of Curves, Proposition 3.49.
Let $B$ be a smooth curve, $0$ a point of $B$ and $B^{\ast} = B \setminus \{ 0 \}$. Let $X \to B^{\ast}$ be a flat family of nodal curves of genus $g$, $\psi:X \to Z$ any morphism to a projective scheme $Z$, ...
Then there exists a branched cover $B' \to B$ and a [projective] family $X' \to B'$ of nodal curves extending the fiber product $X \times_{B^{\ast}} B'$ with the following properties:
The total space $X'$ is smooth.
The morphism $X \times_{B^{\ast}} B' \to X \overset{\psi}{\longrightarrow} Z$ extends to a morphism on all of $X'$. ...
The word "projective" in square brackets in the second line is my addition but was clearly intended, since otherwise we don't have to fill in the central fiber at all. The ellipses conceal conditions concerning marked points, which we won't need.
Let $\bar{X} \to B$ be a flat projective family, whose fibers over $B^{\ast}$ are reduced and nodal. I will take $Z=\bar{X}$ with $\psi$ the inclusion $X \hookrightarrow \bar{X}$. So I get to conclude that there is a branched cover $B' \to B$, and a nodal family $X'$ over $B'$, so that there is a map $\phi: X' \to \bar{X}$. This theorem is stronger than stable and semi-stable reduction in that the new family maps to the old family, but weaker in that we have less control over the fibers of $X'$; they are reduced curves with nodal singularities, but there may be rational components with only one node.
In particular, in your setting, we have a map $\phi: X'_0 \to \bar{X}_0$. Since every component of $X'_0$ is genus $0$, if $\bar{X}_0$ has higher genus, the map $\phi$ is constant. So $\phi(X')$ only meets the central fiber of $\bar{X}$ at only one point. This is absurd, the next paragraph gives one way to turn it into a contradiction.
But $X' \to B'$ is projective, hence proper, and $B' \to B$ is finite, hence proper, so $X' \to B$ is proper. So $X' \times_B \bar{X} \to \bar{X}$ is a closed map. But $X' \times_B \bar{X} = X'$ and the map $X' \times_B \bar{X} \to \bar{X}$ is just $\phi: X' \to \bar{X}$, so we deduce that $\phi$ is a closed map and $\phi(X')$ is closed. As $\phi(X')$ contains all of $X \times_{B'} B^{\ast}$, the map $\phi$ must be surjective, contradiction.