If the collisions took place in space, or on a frictionless road, they would be identical in effect. Whatever reference frame the collision is observed from, all observers would agree about the relative velocities before and after, and the amount of damage done, although they would disagree about absolute velocities.
Suppose the masses are $m$ and $M$ and the initial velocities $v$ and $0$ in case 1, and $2v$ and $v$ in case 2. Suppose also that in both cases the car and truck become attached by the collision, so they move as a single object afterwards. Applying conservation of momentum, the final velocities are $\frac{m}{m+M}v$ and $\frac{2m+M}{m+M}v$ respectively.
The amounts of KE dissipated are :
case 1 : $\frac12mv^2-\frac12(m+M)(\frac{m}{m+M})^2v^2=\frac12 \frac{mM}{m+M}v^2$
case 2 : $\frac12 m(2v)^2+\frac12 Mv^2 - \frac12 (m+M)(\frac{2m+M}{m+M})^2v^2=\frac12\frac{mM}{m+M}v^2 $
Exactly the same in both cases. (The same result - ie no difference in the amount of KE lost - occurs also if the vehicles move apart after the collision.)
If the collision takes place on a road which can supply friction, the only difference is that the vehicles can apply their brakes. If the stationary truck has its brakes on, it is more difficult to move during the collision, the collision time is shorter and the impact forces are greater. In this case the damage will be greater. If neither truck applies its brakes, the collision is the same in both cases. Both vehicles can keep moving after the collision, so not all of the kinetic energy is dissipated.