Wedge constrained motion

In your sketch, the horizontal velocity for block B assumes it is moving on a horizontal surface. For block A to have a vertical velocity, there must be a horizontal force that prevents it from moving to the left. (The normal force has a horizontal component.) The forces must be equal and opposite at the contact surface. The only constraint on velocities is that the two surfaces must remain in contact.

Short answer: It has not so much to do with the 'component along the normal', as it is with the component of velocity 'perpendicular to the surface'.

Imagine any two blocks constrained to maintain contact on one of their surfaces. What should be the minimum necessary condition regarding the velocities of these blocks required to persist this kind of motion? Think.

(HINT: It has something to do with the separation between the contact surfaces of the two blocks.)

Obviously, the first thing that comes to mind is that the separation between the surfaces cannot change, no matter what. It has to remain at zero, or constrained motion will no longer persist; and for the separation between the two surfaces to remain at zero, their relative displacements must be zero as well. As logic follows, their relative velocities must also be zero at all times. This is exactly what we do when we equate their velocities perpendicular to the surface: $(vec v_A(_bot))-(vec v_B(_bot))=0$

Therefore it simply makes intuitive sense to take the components of velocities perpendicular to the surfaces, which also happens to be the direction along which the normal forces act.