Is the normal force in the $x$-direction equal to $mg$ for a sliding object? 
At point A, is the normal force in the x direction equal to mg? 
When the object is going down I imagine $N_x$ decreasing and $N_y$ increasing until point B where $N_x = 0$ and $N_y = mg$. That's why it makes sense that the opposite is true at point A. 
 A: You are working under the incorrect assumption that the normal force needs to have a constant magnitude during the entire motion of the object down the bowl. This is not the case. It is also not the case that $N_y=mg$ at the bottom of the bowl. This is because the object is moving in a circular motion, so the acceleration is upwards at the bottom of the bowl leading to $N_y>mg$.
The normal force is a tricky force to deal with in general. It is essentially the force needed to constrain the object to move along the surface it needs to move along. In general (and certainly in this case) it is not as simple as just looking at a snap-shot of the object at some location and then knowing what the normal force is doing. In many examples (including this one) the normal force will actually depend on the velocity of the object as well as its position. However, we do know that this force always acts perpendicular to the surface (which is why it is the normal force).
So, the best thing to do is include the normal force, and then use some other information to determine what the acceleration is doing that tells us something about the normal force. For example, let's look at point A. Since the normal force is perpendicular to the bowl we already know $N_y=0$ (as you pointed out). However, $N_x$ is different. It depends on the initial velocity of the object. If the object is released from rest, then the object won't be pushing into the bowl at all, so $N_x$ must be $0$. On the other hand, let's say the bowl is a spherical shell and the object was already moving in a circle around this shell. Well then when it gets to point A then $N_x\neq0$ because we need an acceleration towards the center of the shell in order to have circular motion.
The same thing happens at point B. $N_x=0$ because our normal force is perpendicular to the bowl. If the object is just sitting at point $B$ then $N_y=mg$ since there is no acceleration in the vertical direction. If the object has just slid down the bowl then $N_y>mg$ because our acceleration is now pointing upwards at point B to allow for circular motion.
TL;DR The normal force doesn't just depend on position, and it certainly doesn't have the same magnitude during the entire motion of the object. Therefore, the logic of the components "switching values" is unwarranted.
A: A normal force is an opposing force that acts between two surfaces in contact.When object is on a horizontal ground and gravitational force is acting ,because of 0 acceleration we assume there a opposing normal force to the gravitational force to balance it.In vertical surface like this the gravitational force is acting in vertical direction and no other force is acting to the rock towards wall,so the wall will not provide any normal force.If we take mg force by wall of bowl  in x direction instead of 0,the rock should accelerate in x direction.So there is no normal force by wall in x direction.
At positions from A to B(If the object was not in motion)the normal force to the object by wall of bowl is increasing from 0 to mg(gravitational force).In case the object is in motion from point A to B ,centripetal force will also act on object towards wall.In that case normal force has to balance gravitational force and centripetal force.
A: HINT:  Consider the following argument about an object sitting at rest on a table:


*

*There is no acceleration in the vertical direction.  

*This means that there is no net force on the object in the vertical direction.  

*This means that the vertical component of the normal force $N_y$ must cancel out the object's weight $mg$.  

*Thus, $N_y = mg$.


This argument fails for your problem.  Where does the argument fail?
A: I can provide a bunch of hints, and that might help


*

*At point A the contact force is zero( think of drawing a tangent to the the hemisphere at that point, you'll see it's parallel to gravitational force on particle and since this tangent represents the instantaneous surface of contact you can see the rock isn't really pushing on the surface and so contact force is zero.

*Contact force in this case is a constraint force and acts always perpendicular to the trajectory of the particle and hence it does no work.

*Contact force at any point is perpendicular to the tangent you draw there.

*At point B there is a certain centripetal force involved so contact force won't be equal to mg as @garyp pointed in the comments.

A: The normal force is perpendicular to the surface; that's it. At point A, it's directed towards the center of the circle, which is what I think you're calling "the x-direction". At point B, it's directed towards the center of the circle, which is what I think you're calling "the y-direction".
In order to determine the magnitude of the normal force, you need to solve Newton's second law. If you did so, you would find that the normal force is zero at point A (no initial speed, no centripetal acceleration, no net force in the horizontal direction), while at point B, $F_N>mg$ (since there is some centripetal acceleration, directed upwards, so the normal force must be greater then the gravitational force).
