I am going to try this without drawing anything out. Let us use the example of 2-methylcyclohexanone and methyl vinyl ketone. If the aldol preceded the Michael addition, then you must perform the equivalent of the Michael addition (or a new mechanism) to complete the reaction. I think if you draw that aldol product, two challenges should be found. One, the hydrogens are slightly less acidic because they are now delta to the carbonyl group. Secondly, once enolization takes place, the carbonyl oxygen accepting electrons will have a negative charge in the greater resonance contributor.
If the conjugate addition takes place first, then the aldol condensation can take place normally between the two ketones.