Making this a bit more complex:

Imagine instead of the cars crashing, one car throws (throwing without extra force) a stone of 1 kg at the other car.

The kinetic energy it gets from that car is 0,5*1*10^2 = 50. The kinetic energy the other car contributes is, according to the above, another 50.

Why another 50? If a stone of mass 1 kg travelling at +10 m/s collides with a car of mass 1000 kg with a velocity of -10 m/s, the total kinetic energy of the collision is

\(\displaystyle KE = KE_{stone} + KE_{car} = 50 + 50000 = 50050 J\)

I don't think this number is very useful though.

No the energy with which the stone hits the other car is 100. So the relative speed of the stone between the two cars is 20 m/s, but the energy with which it impacts is equal to only (100 = 0,5*1*v^2; v=) 14,14 m/s. Is that correct?

Nope.

If you're driving a car and someone throws the stone at your car, then in your reference frame, you will see a stone travelling towards you at 20 m/s. If, after the collision, the stone is inside your car, it is travelling now with you in your car, so the total energy exchanged is

\(\displaystyle \Delta E = \frac{1}{2} \cdot 1 \cdot (20)^2 = 200 J.\)

It is often easier and more enlightening in collision problems to consider momentum exchange. In this case you have a stone with momentum

\(\displaystyle p_1 = mv_1 = 1 \cdot 10 = 10\) kg m/s

Colliding with a car. After the collision, the momentum of the stone is

\(\displaystyle p_2 = mv_2 = 1 \cdot -10 = -10\) kg m/s

So the total momentum exchange, \(\displaystyle \Delta p = p_1 - p_2\), from the stone to the car is 20 kg m/s. If this collision occurs over 0.01 seconds, then the force is

\(\displaystyle F = \frac{\Delta p}{\Delta t} = \frac{20}{0.01} = 2000 N\)

And the energy transferred from the stone to the car can also be calculated from the momentum exchange:

\(\displaystyle \Delta E = \frac{\Delta p^2}{2m} = \frac{20^2}{2 \cdot 1} = 200 J\)