It will never be surjective if $\bar{K} \ne K$ and $X$ is of positive genus. If $K$ is a finitely generated field, the Picard group over $K$ is a finitely generated abelian group (Mordell-Weil), while over the algebraic closure, it's definitely not finitely generated, it has both infinite torsion and "infinite rank". If $K = \mathbb{R}$, then the Picard group is a $g$-dimensional real manifold and, over $\mathbb{C}$ is a $g$-dimensional complex manifold.
For a positive dimensional algebraic variety $V/K$, if $K \ne \bar{K}$, then $V(K) \ne V(\bar{K})$. So this has nothing to do with Picard groups. Proof: WLOG take $V$ to be a plane curve $f(x,y)=0$ and it has points with the $x$-coordinate taken from $\bar{K} \setminus K$.