I've read through the past posts on this topic and still am not finding an answer to this specific aspect. I get that in the "no friction" case, gravity pulls the car down on the track and the resulting Normal force provides both the centripetal force and the "y direction" force that counteracts gravity (which keeps car from sliding down track). What I don't get is what happens as car speeds up. Does the y-component of the normal force remain "mg" and just the x-component of the normal force increases as a reaction force to the car speeding up (and so, colloquially, wanting more and more to go in a straight line and so banked track "hits" it harder as it turns it). Or does the normal force in the direction perpendicular to the track increase as a reaction force to increased speed (and so the normal force in both the x and the y-direction bothincrease?)
A "normal" force is normal or orthogonal to the surface creating the force. In other words, the ground or track can only push straight out from the surface. (Forces along the surface would be due to friction, but you are not considering friction here).
Since the direction of the force is constant, that means you cannot change the $x$ component of the force and leave the $y$ component unchanged when both are non-zero. Increasing one requires increasing the other.
If increased speed increases the centripetal force, then it increases the upward force as well.