This is ok when $X$ is smooth and $Y$ and $Z$ are integral. The point is that over a smooth curve, any finite morphism from an integral curve is automatically flat: over a Dedekind scheme, flat is the same as torsion-free [Tag 0AUW], and a finite integral extension is indeed torsion-free.
Then $Y \times_X Z \to X$ is also finite flat, so satisfies both going up [Tag 00GU] and going down [00HS]. This implies that all irreducible components of $Y \times_X Z$ have dimension $1$, and for any point $p \in Y \times_X Z$ the height of $p$ is the same as the height of its image in $X$. In particular, every generic point of an irreducible component maps to the generic point of $X$, so there is only one irreducible component if and only if $K(Y) \otimes_{K(X)} K(Z)$ has only one irreducible component.
Finally we have to say something about reducedness. Recall that a scheme $S$ is reduced if and only if it is (R$_0$) and (S$_1$) [Tag 031R]. For $Y \times_X Z$, the condition (R$_0$) means exactly that $K(Y) \otimes_{K(X)} K(Z)$ is reduced. Thus, we see that $K(Y) \otimes_{K(X)} K(Z)$ is a field if and only if $Y \times_X Z$ is (R$_0$) and irreducible.
Recall also that a $d$-dimensional Noetherian scheme $S$ is Cohen–Macaulay if and only if it is (S$_d$), hence a $1$-dimensional scheme $S$ is (S$_1$) if and only if it is Cohen–Macaulay. So it remains to show that $Y \times_X Z$ is always Cohen–Macaulay. But $\Delta_X \colon X \hookrightarrow X \times X$ is a regular closed immersion [Tag 0E9J], hence so is $Y \times_X Z \hookrightarrow Y \times Z$ since it is the pullback of $\Delta_X$ along $Y \times Z \to X \times X$ [Tag 067P]. Since $Y$ and $Z$ are reduced curves, they are (S$_1$), hence Cohen–Macaulay as they have dimension $1$. Then $Y \times Z$ is Cohen–Macaulay as well [Tag 045J], hence so is $Y \times_X Z$ since it is a regular closed subscheme of $Y \times Z$ [Tag 02JN].
