Well Patience...
you're trying to understand the concept of transmission from an outside observer point of reference (aka car passing by)
That is quite challenging if you do not fully understand the much simpler point of reference first
(be the car !)
So if you move the frame of reference to the car itself, the observation is very simple indeed.
linear motion of the pistons is transfered to rotation via the crankshaft, which is then passed along through the clutch to various gears (no matter if right angled as for a driveshaft or not)
to the rear wheels.
And the wheels then push the road in a specified direction.
See.. very simple for this frame of reference.
Now the outside observer has again the wheel pushing the -now fixed- road and thus the wheels push the axle, which in turn push their bearings, which in turn push their bearing retainer, which push... endless number of nuts and bolts... which then push the chassis which in turn takes all the rest of the parts with it.
But frankly you shouldn't overcomplicate things at all...
in physics all frames of reference are equally valid,
and picking the easiest one first is by far the most convenient way of doing this
And since we mainly talk about friction when it comes to moving cars(bikes/karts/trolleys...)
we can safely assume that a structurally sound force fitted connection (nuts/bolts) will hold up much longer than a non permanent friction fitted connection (tyre/road).
So yeah, why bother with that.
If you want to know if a car itself is being pushed or pulled by the wheels, then you must differentiate between RWD and FWD; since that would be determined by thecenter of mass of the vehicle in regards to the driven axle..(beware of All wheel drives

a huge PITA to calculate just assume it pushes AND pulls

)
If however you want to know the forces applied to a certain part of the car/kart; then
it's fairly easy again..
say the rear axle bearing..
take the easiest one (say gravity) add a second force where applicable (forward momentum)
calculate the resulting vector, add the exact same forces and add them to the outer bearing race on the opposite sides so the two cancel out.. et voila; you're done.
(see, the easiest frame of reference

)
'sid