I try to understand the proof of an extension theorem proved in Moonen's and van der Geer's draft Abelian varieties(in online accessible notes Theorem 1.18 on page 14):
(1.18) Theorem.
Let $X$ be an abelian variety over a field $k$. If $V$ is a smooth $k$-variety then any rational map $f: V \dashrightarrow X$ extends to a morphism $V \to X$.
Proof:
We may assume that $k = \bar{k}$, for if a morphism $V_{\overline{k}}= V \times_k \overline{k} \to X_{\overline{k}}$ is defined over $k$ on some dense open subset of $V_{\overline{k}}$, then it is defined over $k$. Let $U \subset V$ be the maximal open subset on which $f$ is defined. Our goal is to show that $U = V$. [...]
Questions: the assumption $k = \bar{k}$ I not understand. if we succeed to prove that $V_{\overline{k}} \to X_{\overline{k}}$ extends to a morphism, why $f$ is also extendable? Afterwards, if we proof $V_{\overline{k}} \to X_{\overline{k}}$ extends to a morphism, why if it is defined over $k$ on some dense open subset of $V_{\overline{k}}$ (=preimage of a dense open $U \subset V$ under projection $V_{\overline{k}} \to V$), then it is defined over $k$?