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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.