I think I see the connection now.
Let $D\in\operatorname{Div}(X)$ be any divisor such that $DF=1$. We'll show that $D\sim S+nF$ for some section $S$ and integer $n\in\mathbb Z$. For each $n\in\mathbb Z$, let $D_n=D+nF$, and let $K$ be a canonical divisor on $X$. Since $F$ is a fiber, $F^2=0$ and the adjunction formula then shows that $KF=-2$. Given this, we have the equalities
\begin{align*}
D_n^2=D^2+2n && D_nF = 1 && D_nK=DK-2n
\end{align*}
Using Riemann-Roch, we now see that
$$h^0(D_n)+h^0(K-D_n)\ge\frac12D_n(D_n-K)+\chi(\mathscr O_X)=\chi(\mathscr O_X(D))+2n.$$
Furthermore, we have $(K-D_n)F=-3<0$, so $h^0(K-D_n)=0$. Hence, for $n$ large enough, we have $h^0(D_n)>0$ and so there's some effective divisor $E\sim D_n$. Now, because $E$ is effective and because $EF=1$, it must be of the form $E=S+mF$ where $S$ is a section and $m\ge0$. Hence $D\sim S+(m-n)F$ as desired.
The above shows that there's some divisor of the form $D=S+nF$ for which $\mathscr O_X(D)\simeq\mathscr O_X(1)$. Hence,
$$\mathscr O_X(1)\cdot\pi^*\bigwedge^2E=\deg(\pi^*E\vert_S)+n\pi^*\left(\bigwedge^2E\cdot c\right)=\deg(\pi^*E\vert_S)=\deg(E),$$
where $c\in C$ is just some point. The second to last equality comes from the fact that, as a cohomology class, $\bigwedge^2E\cdot c$ lives in $\operatorname H^4(C,\mathbb Z)=0$, and the last equality comes from the fact that $\pi\vert_S:S\to C$ is an isomorphism sending $\pi^*E$ to $E$.
Edit: The above argument is essentially correct, but contains a slight error. I am treating all fibers as equivalent divisors, but this is not quite the case. All fibers (or rather, their first Chern classes) are equivalent in $H^2(X;\mathbb Z)$ (which is all that matters of computing intersection numbers), but they are not necessarily equivalent in $\mathrm{Pic}(X)$. The only change this has on the argument is that you fix a particular fiber $F$, consider these same $D_n$'s, and then when you get an effective $E\sim D_n$ (for $n\gg0$), you cannot conclude that $E$ is of the form $S+nF$; instead, you know it is of the form $S+\pi^*P$ where $P\in\mathrm{Div}(C)$ is some divisor on $C$. This is still good enough for the short argument in the last paragraph.
