I have a pretty basic pulley problem where I lack the right start.
A child sits on a seat which is held by a rope going to a cable roll (attached to a tree) and back into the kid's hands.

When it sits still, I believe that the force on either side of the rope must be equal to keep is static, therefore each rope holds $\frac{1}{2}mg$, the cable roll has to carry the full $mg$.
Now, the kid wants go up with $\frac{1}{5}g$. For the whole system to accelerate up, the cable roll has to support another $\frac{1}{5}mg$ resulting in $\frac{6}{5}mg$ of force.
The question that I cannot answer is:
How much force does the kid need to apply onto the rope in its hands?
As I said before, $F_k$ (kid) and $F_s$ (seat) have to be $\frac{6}{5}mg$. So I get this:
$$F_k + F_s = \frac{6}{5}mg $$
In order to solve for either one, I would need another equation. The forces cannot be equal, otherwise there would be no movement of the rope. So I just invented the condition, that the difference of the forces has to be the acceleration:
$$F_k - F_s = \frac{1}{5}mg $$
I can solve this giving me $F_s = \frac{5}{10}mg$ and $F_k = \frac{7}{10}mg$ which will sum up to the total force.
But is this the right approach at all?