You can compose the projection $D^b(X) \to A$ with the equivalence $A \cong D$ and the embedding $D \to D^b(Y)$. This will be compatible with one of the equivalences. Or you can consider an analogous composition $D^b(X) \to B \to C \to D^b(Y)$. Moreover, you can take the direct sum of these two functors. This will be compatible with both equivalences. But if you want to get more interesting functor you will need some compatibility between the gluing data of the decompositions.

You can compose the projection $D^b(X) \to A$ with the equivalence $A \cong D$ and the embedding $D \to D^b(Y)$. This will be compatible with one of the equivalences. Or you can consider an analogous composition $D^b(X) \to B \to C \to D^b(Y)$. Moreover, you can take the direct sum of these two functors. This will be compatible with both equivalences. But if you want to get more interesting functor you will need some compatibility between the gluing data of the decompositions.

EDIT. Let $X = Y = P^2$ and consider the semiorthogonal decomposition $$ D^b(P^2) = \langle O,O(1),O(2) \rangle. $$ Let us construct a FM-functor interchanging the $O(1)$ and $O(2)$ components. First, the projection to $O(2)$ is given by $RHom(O(2),-)$. The FM-functor which projects to $O(2)$ and then embbeds into $O(1)$ is thus given by the object $$ O(-2)\boxtimes O(1) $$ on $P^2\times P^2$. Further, the projection to $O(1)$ by $RHom(T(1)[-1],-)$, so the FM-functor which projects to $O(1)$ and then embbeds into $O(2)$ is given by the object $$ \Omega(-1)\boxtimes O(2)[1]. $$ So, in the end you can take the direct sum of these two objects as a kernel. This will give you a FM-functor compatible with your equivalences.