There is no change in rotational or linear velocity during the process.
Can you explain the work done by the person rolling the ball, in the two cases, ideally. I would prefer an explanation without using conservation of energy theorem, but that is also fine. I am getting confused in both my approaches.
A ball on a hill is pulled on by gravity, making a force (parallel to the ground, pointing down the hill) of $F=mg \sin(\theta)$. Acting at its center-of-gravity.
To roll a ball up a hill requires a force of half that much, if applied at the top of the ball, half because of leverage if you push on the top. The distance we need to apply this force is twice the length of the hill. Work done is $$Fd= (\tfrac{1}{2}mg \sin(\theta))~(2L)= mgL \sin(\theta)$$ The height of the hill is $$h= L \sin(\theta) \implies \text{Work}=mgh$$
So we expend work energy of $mgh$ and increase the ball’s height potential energy by $mgh$. But the leverage let us use half the force over twice the distance. Assumes no friction. Wasn’t sure exactly what you were looking for maybe that helps.
A person rolling a ball up hill does positive work to increase the potential energy and overcome friction. The force is in the direction of motion. To keep the ball moving at a constant speed coming down the hill, he exerts a force opposite to the direction of motion and does negative work. He may have help from friction in preventing an increase in kinetic energy.
