What force causes the ball to move up?
Floris has given you a correct answer but he hasn't addressed the misconception that is preventing you from seeing it as correct.
In the question you write
In position 2, [...] (because gravity force is greater than ground reaction force acting on the ball).
and in the comments on Floris' answer similarly
the normal force (the force exerted on the ball by surface it's situated on) is 'at most' equal to $mg$ - its weight (it's obviously smaller when the surface isn't parallel to the ground)
but these statements are wrong.
The key is understanding the normal (or solid reaction) force better.
In the simplest first-course understanding the normal force has two properties
- perpendicular to the region of contact and
- takes on whatever magnitude is required to keep the objects from inter-pentrating.
You seem to be very clear on (1) but have misunderstood (2).
So, what happens when the ball reaches the point where the slope begins? Well, the normal force begins to point a little bit backward instead of completely up due to property (1), but it also grows to have larger magnitude because of property (2) and the fact that the only way for the ball to continue forward without interpenetration is to go upward as well.
The result then is a net force on the ball that is upward and backward, causing the ball to rise and slow.
Some interesting question to ask yourself are:
What happens to the normal force if the slope becomes constant instead of continuing to change?
What happens to the normal force as the ball slows toward a stop at it's maximum height? What value does it have at the maximum height?
What happens to the normal force as the ball descends the ramp again?
The answers are slightly subtle, which is one of several reasons we don't do changing slopes in the introductory course.
You answered this question yourself. You state: "In position 1, the gravity force acting on the ball is cancelled out by the ground reaction force, so the net force is zero.". The ground reaction force you mentioned keeps the ball from falling through the surface. It can be considered as the repulsion from the electron cloud surrounding the ball by the electron cloud of the surface. As the ball reaches the slope, the direction of this repulsive force changes from perfectly vertical to slightly tilted, causing a horizontal force that will slow the ball down. Due to the velocity of the ball hitting the slope the electron cloud are pushed closer, increasing the upward force (I guess).
There is a normal force from the surface on the ball. As long as the surface is straight, this force is equal to the force of gravity (the weight of the ball). Once you reach the curve, the normal force increases and the ball is accelerated upwards. This is the same thing that happens when a ball follows a circular track: there is an additional centripetal force that changes the direction of the ball (in the case of a circular track, the change is perpendicular to the velocity and the total speed doesn't change).
In this case, as the velocity changes, there will appear a component of gravity along the surface of the track - this slows the ball down. Your premise that the horizontal component of the velocity of the ball doesn't change ("at constant horizontal velocity $v$") is wrong unless there is another external force to offset the pull of gravity.