See Theorem 4.2.2 in https://www-fourier.ujf-grenoble.fr/~mbrion/lin.pdf

In particular, in your example properness of $X=G/B$ simplifies the left part of the sequence, turning it into $$0\to \hat{G}\xrightarrow{\gamma} Pic_G(X)\to Pic(X)\to Pic(G\times X)$$ where $\hat{G}$ is the group of characters of $G$ and $\gamma$ maps $\chi$ to $X\times_G \chi$. The map $Pic_G(X)\to Pic(X)$ is just the forgetting map, for the definition of the last one, see the text.

Also, if $G$ is affine, as in your case, $Pic(G\times X)=Pic(X)\times Pic(G)$(I write $=$ instead of $\cong$ to emphasise that the isomorphism is given by projection maps) so, applying Proposition 4.2.3, we get $$0\to \hat{G}\to Pic_G(X)\to Pic(X)\to Pic(G)$$

See Theorem 4.2.2 in https://www-fourier.ujf-grenoble.fr/~mbrion/lin.pdf

In particular, in your example properness of $X=G/B$ simplifies the left part of the sequence, turning it into $$0\to \hat{G}\xrightarrow{\gamma} Pic_G(X)\to Pic(X)\to Pic(G\times X)$$ where $\hat{G}$ is the group of characters of $G$ and $\gamma$ maps $\chi$ to $X\times_G \chi$. The map $Pic_G(X)\to Pic(X)$ is just the forgetting map, for the definition of the last one, see the text.

Also, if $G$ is affine, as in your case, $Pic(G\times X)=Pic(X)\times Pic(G)$(I write $=$ instead of $\cong$ to emphasise that the isomorphism is given by projection maps) so, applying Proposition 4.2.3, we get $$0\to \hat{G}\to Pic_G(X)\to Pic(X)\to Pic(G)$$

Edit: As Marc Hoyois mentions, if $X$ is a $G$-bundle over $Y$, then $Pic_G(X)=Pic(Y)$, see lemma 3.3.1