Let $\pi:X\to\Bbb{P}^1$ be a rational elliptic over an algebraically closed field $k$. If $K:=k(t)$, we have the Mordell-Weil group $E(K)$, which is an elliptic curve over $K$ (which is also the generic fiber of $\pi$).
I'm trying to understand some consequences of the well-known isomorphism $E(K)\simeq \text{NS}(X)/T$.
Considering the natural correspondence between points $P\in E(K)$ on the generic fiber with divisors $(P)$ on $X$, the map $P\mapsto (P)\text{ mod }T$ is natural.
For the inverse map $\psi:\text{NS}(X)/T\to E(K)$, Schuett and Shioda explain it like this (I'll use some language abuse).
If $D\in\text{Pic}(X)$, we write $D=D'+D''$ where $D'$ is the horizontal part and $D''$ is the vertical part. The horizontal part $D'$ intersects $E$ properly with intersection number $D'\cdot E$. We then are able to find $P\in E(K)$ such that $D'-(D'\cdot E)$ is linearly equivalent to $P-O$ in $E$. Then we let $\psi(D):=P$.
My question is this: suppose we have sections $P_1,P_2,P_3\in E(K)$ and the intersection numbers $P_1\cdot P_3=a$ and $P_2\cdot P_3=b$. We then have $((P_1)+(P_2))\cdot P_3=a+b$ (I emphasize that the sum $(P_1)+(P_2)$ is a sum of divisors on $X$).
By the Shioda-Tate isomorphism, can I say that there is some $P_4\in E(K)$ such that $P_4$ equivalent to $(P_1)+(P_2)$ and therefore $P_4\cdot P_3=a+b$? If I understand correctly, I can, since clearly $D:=(P_1)+(P_2)$ is already horizontal.
I'm I getting something wrong here?