Let $K$ be a field of characteristic zero and $X_K$ a smooth projective curve on $K$. Denote by $\bar{K}$ the algebraic closure of $K$ and $X_{\bar{K}}$ the base change of $X_K$ to $\bar{K}$. Under the natural morphism from $X_{\bar{K}} \to X_K$, we have an induced morphism $\mbox{Pic}^0(X_K) \to \mbox{Pic}^0(X_{\bar{K}})$. Is this morphism surjective? If not true in general, is there any known condition on $K$ under which this holds true?

$\begingroup$ Are you asking if the morphism of schemes is surjective? Or the morphism of groups? $\endgroup$ – Daniel Litt Sep 6 '14 at 1:19
First of all, the image of your homomorphism is invariant under the Galois group $G:=\mathrm{Gal}(\bar{K}/K)$. So the right question is to ask whether the induced homomorphism $\mathrm{Pic}(X_{K})\rightarrow \mathrm{Pic}(X_{\bar{K}})^G$ is surjective.
In our answer to this question of yours, we (Daniel Loughran and me) give elements for the answer: the HochschildSerre spectral sequence gives an exact sequence $$0\rightarrow \mathrm{Pic}(X_K)\rightarrow \mathrm{Pic}(X_{\bar{K}})^G\rightarrow H^2(G, \bar{K}^*)\rightarrow H^2(X_{K},\mathbb{G}_m)$$ $H^2(G, \bar{K}^*)$ is the Brauer group $\mathrm{Br(K)}$ of $K$. So a sufficient condition is that $\mathrm{Br(K)}=0$. This happens for instance if $K$ is the field of rational functions of a curve over $\mathbb{C}$, or the maximal unramified extension of $\mathbb{Q}_p$, or an algebraic extension of $\mathbb{Q}$ containing all roots of unity  see Serre, Local fields, ch. X.
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 (MordellWeil), 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 $\bar{K} \setminus K$.