Let's say Jeff can pull a maximum of F newtons force pulling or pushing.
In method A, the maxinum pulling force on the rope will be F newtons.
In method B, the pulling force on the rope is much more than F newtons.
The tension on the rope is the pulling force on the rope, say T newtons. The F pushing the rope at about the rope's midpoint will impart tension T on the rope.
Get at the intersection of the F and the two opposing T's where the rope is being pushed. One T points to the tree; the other T points to the car.
The components (we call Tp) of these two T's that are parallel to the F will equal or neutralize the F. So,
2(Tp) = F
Tp = F/2 = (0.5)F newtons.
The components (we call Tr) of these two opposing T's perpendicular to F will neutralize each other. [You thought of them as "dissipated forces".]
Let us say the rope is bent 178 degrees at the point of pushing. Then a T is 89 degrees from the direction of F. Hence,
T*cos(89deg) = 0.5 F
T = (0.5 F) / cos(89deg)
T = (28.65)F newtons -----which is a lot more than F newtons.