Consider a geometrically ruled surface $\pi: S\longrightarrow C$ where both $S$ and $C$ are projective, non-singular and complex ($C$ is obviously a curve). By using Tsen theorem one can show that there exists a section $\sigma:C\longrightarrow S$ such that $\pi\circ\sigma=id_C$.
Now, for every fiber $F$ of $\pi$, it is clear that $\sigma(C)\cap F$ is a single point.
But why $F.\sigma(C)=1$?
I mean, a fibre $F\cong\mathbb P^1$ and $\sigma(C)$ could have the same tangent line on the intersection point.
(This ''fact'' is assumed as true without any explaination in Hartshorne book in the proof of the proposition $V.2.2$)
No, a section cannot be tangent to a fibre at any point. $\pi \circ \sigma = \operatorname{id}$ implies that at any point $p = \sigma(q)$ in the image $\Gamma$ of the section, $(d_p \pi)_{|T_p \Gamma}$ must be an isomorphism onto $T_q C$.
On the other hand, $(d_p \pi)_{|T_p F} =0$ because $\pi$ is constant along $F$.
So if $\Gamma$ and $F$ share a tangent vector at $p$, then $(d_p \pi)_{|T_p \Gamma}$ is not injective, a contradiction.