The centre of mass is irrelevant except for calculating the force of gravity.

For the car to move in the direction shown with the green arrow in your plot, the ground is pushed by a frictional force in the opposite direction. As incremental (v a vector) momentum of car is $mv$, the moon is pushed back by incremental -2 $mv$ , for conservation of momentum , this should equal $Mv^{'}$, M the mass of the moon and $v^{'}$ its velocity from momentum conservation. The mass of the moon is practically infinite with respect to the mass of the car, so the velocity of the moon would be very small, so nothing really moves except the car.

If the moon had the same mass as the moving object, this would result in a rotation of the moon about the center of mass.

The centre of mass of the moon is irrelevant except for calculating the force of gravity. It is the total center of mass, car+moon, that is important in calculating kinematic variables for a system of two bodies. The following simple argument may help.

For the car to move in the direction shown with the green arrow in your plot, the ground is pushed by a frictional force in the opposite direction. As incremental (v a vector) momentum of car is $mv$, the moon is pushed back by incremental -2 $mv$ , for conservation of momentum , this should equal $Mv^{'}$, M the mass of the moon and $v^{'}$ its velocity from momentum conservation. The mass of the moon is practically infinite with respect to the mass of the car, so the velocity of the moon would be very small, so nothing really moves except the car.

If the moon had the same mass as the moving object, this would result in a rotation of the moon about the center of mass.