Vakhil 14.2.E why $t \mapsto t, \mathcal{O}_X(D) \to \mathcal{O}_X$ is isomorphism?

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I found that question on MSE. And have misunderstanding. Let $X = \mathbb{P}^1$, $p \in \mathbb{P}^1$ - point, $L=\mathcal{O}_X(p)$.

Then $L_p = \{\frac{p(t)}{t} | p(t) \in k[t] \}$ and $\mathcal{O}_{X,p} = \{p(t) | p(t) \in k[t] \}$ where $t$ is local parameter in point $p$. So the formula $f \mapsto f, L_p \to \mathcal{O}_{X,p}$ is not even define morphism correctly (where we must send $\frac{1}{t}$?). But there is in related SE question:

Part of the exercise to show that $B = \{open\: U\subset X:\: O(div(s)) \rvert_U \approx O_U \}$ is a base for the Zariski topology on X. On each such set it is easy to see that $O_X(D)(U)\longrightarrow O_X(U)\longrightarrow L(U)$ defined by $t\longmapsto t\longmapsto ts$ is an isomorphism, so in the end we find that $O_X(div (s))\approx L$ .

Hope somebody helps me with my misunderstanding.