$F_2\circ F_1\dashv G_1\circ G_2$ and $F_2\dashv G_2$ but not $F_1\dashv G_1$
Solution 1:
Your example is correct. To see this, first note that the inclusion of $\mathbf{Sheaf}$ into $\mathbf{separatedPresheaf}$ has a left adjoint given by the sheafification functor (this follows by the universal property of sheafification). It then remains to show that on separated presheaves, the functor $(-)^+$ agrees with the sheafification, or in other words that $F^+ \simeq F^{++}$ for every separated presheaf $F$. But this follows from $F^+$ being a sheaf in this case.