The problem with attempting to do the analysis with the forward point of contact on the box when it is sliding is that the box is accelerating. This makes a non-inertial frame and there's more moving parts. Besides the force of gravity on the center of mass, there will be fictitious forces.
First, lets assume friction is zero. If so, we can calculate the acceleration of the box.
$$F_{x'} = mg\sin \theta$$
$$a_{x'} = g\sin \theta$$
In an accelerating frame, we get fictitious forces.
$$ F_{fict} = -ma_{frame}$$
$$ F_{fict} = -mg\sin \theta$$
The sign indicates the force is opposite the direction of the acceleration. The case where we expect the box to tip most likely is where it is tall and not wide. Let's assume the width is minimal (a rod sliding on its end). If so, the torque from gravity will be the COM at half the height, and leaning forward $\sin \theta$
$$\tau_g = \frac12 hmg \sin \theta$$
Meanwhile the fictitious force is located at half the height and points opposite the acceleration vector, so it has a lever arm of exactly half the height.
$$\tau_{fict} = F_{fict} d$$
$$\tau_{fict} = -\frac12 hmg\sin \theta$$
$$ \tau_{net} = \tau_{g} + \tau_{fict}$$
$$ \tau_{net} = \frac12 hmg \sin \theta - \frac12hmg \sin \theta$$
$$ \tau_{net} = 0$$
If you add friction, you will reduce the acceleration of the box, reducing the fictitious force in this frame. When that happens, the net torque would be sufficient to tip a rod, but would depend on the dimensions if a box would tip over.
I didn't do the math, but $\tau_{fict}$ should be pretty simple to calculate, and then $\tau_{g}$ becomes $\frac12h\sin \theta - \frac12w\cos \theta$ (I think)...
Sorry, I wanted to reply to this comment, and I couldn't do it in the comment space.
If we instead take the static plane to be the frame of reference (and
a minimal width box), wouldn't we get the same results anyways?
The problem is that in the frame of the plane, the axis you want to consider in the question (the front corner of the box) is accelerating. I don't think that's okay to do. Let's imagine a pendulum being accelerated via a force on the axis. And we let it reach steady state so the angle isn't changing.
The solid pendulum has a force accelerating it to the right, and it has gravity pulling it down. The force is applied directly at the axis and contributes no torque. Gravity pulls the weight down, so is applying a torque of $(mg \times L\sin \theta)$. Yet the pendulum does not rotate. Perhaps someone else has more information about this situation. My intuition is that this is simply the wrong technique for an accelerating axis, but I'm not sure what the correct analysis would be.
Why does it accelerate linearly, if the only force I'm applying is a torque?
Forces and torques are separate things, but still related. A torque is produced by a force
$$\mathbf\tau=\mathbf r\times\mathbf F$$
So fundamentally you are just applying a force. That affects the linear motion of the object. If your force also happens to have a torque about the COM of the object, then the object will start spinning as well.
In other words, your force is not a torque. Your force has a torque about the COM. Torque is a manifestation of where and in what direction the force is being applied to the object.
Another way to look at this is to compare units. Forces have units of $\rm N$ where as torques have units of $\rm N\cdot\rm m$, so they cannot be the same thing.
Now, getting to how a tangential force causes a linear acceleration, it is easier to go more general. Let's say we have a rigid body of total mass $M$ consisting of $N$ particles. We know that the center of mass of the body is given by
$$r_{cm}=\sum_i^N\frac{m_i\mathbf r_i}{M}$$
Taking time derivatives, we can write the velocity and acceleration of the center of mass as well
$$v_{cm}=\frac{\text d}{\text dt}r_{cm}=\sum_i^N\frac{m_i\mathbf v_i}{M}$$
$$a_{cm}=\frac{\text d}{\text dt}v_{cm}=\sum_i^N\frac{m_i\mathbf a_i}{M}$$
Using Newton's second law we know that, for each particle
$$\mathbf F_i=m_i\mathbf a_i$$
and so putting it all togther we see that
$$a_{cm}=\sum_i^N\frac{\mathbf F_i}{M}$$
or rewriting things
$$\sum_i^N\mathbf F_i=Ma_{cm}$$
Now we need to think about the force we apply and the forces on each particle. The net force on any particle is the sum of any external forces acting on that particle and any forces between the other particles in the body:
$$\mathbf F_i=\sum_{ext}\mathbf F_i^{ext}+\sum_{j}^N\mathbf F_{j\rightarrow i}$$
Therefore:
$$\sum_i^N\mathbf F_i=\sum_i^N\sum_{ext}\mathbf F_i^{ext}+\sum_i^N\sum_j^N\mathbf F_{j\rightarrow i}$$
Now, this double sum in the second term covers all $(i,j)$ pairs, so we can rewrite this term by double counting and then correcting for our double counting:
$$\sum_i^N\mathbf F_i=\sum_i^N\sum_{ext}\mathbf F_i^{ext}+\frac12\sum_i^N\sum_j^N\mathbf (\mathbf F_{j\rightarrow i}+\mathbf F_{i\rightarrow j})$$
However, we know that by Newton's third law, $\mathbf F_{j\rightarrow i}=-\mathbf F_{i\rightarrow j}$ So our "double count" sum is actually $0$. Therefore:
$$\sum_i^N\mathbf F_i=\sum_i^N\sum_{ext}\mathbf F_i^{ext}=\mathbf F_{net}^{ext}$$
Therefore, bringing everything together:
$$\sum_i^N\mathbf F_i=\mathbf F_{net}^{ext}=Ma_{cm}$$
Notice how we have not assumed anything about the torques of these forces! So we see that, independent of what torques the applied forces might have, external forces alter the linear acceleration of the center of mass of the system of particles.
Best Answer
They don't want to get in front of the rolling wheels -- at least, not intrinsically.
When you are turning, your front wheels stop going forwards so much, and start going sideways. As Newton taught us, a great way to think about such things is that an object in motion will tend to keep moving however it is moving. There are sideways forces on the front wheels created by the friction of the wheels with the ground, slowing down its forward speed, and transferring that into sideways speed. So that's how "objects keep moving" works out for the front side of your car: forces act on it to stop its forward motion and create sideways motion. Now let's examine how this principle works for the backside of your car.
Normally, there is an "easy direction" for the back of your car to move -- in the direction of the wheels -- and a "hard direction" to move -- perpendicular to that direction. This creates side-forces which keep your rear wheels "following" your front wheels. They have a less-steep turn than the front wheels but also they start acting a few meters before the front wheels do, so they tend to take turns tighter, which is why it's harder, e.g., to forwards-perpendicular-park on your right rather than your left (if you've just been succeeding unconsciously or always back-in: when you forwards-perpendicular-park to the right, you usually have to swerve left pre-emptively to give your rear wheels room to take the turn, because taking that turn tighter than the front wheels would otherwise mean hitting the car occupying the spot 1 before the one you want to park in).
When you pull the handbrake, or your rear wheels hit ice, hydroplane, or go airborne, you set these two directions on equal footing: your rear wheels then are not so well "guided" by these side forces, and mostly just want to keep going forward, with a slight pull induced by the front tires. Even worse, as @BillN notes, they now have a lower coefficient of friction than those stabilizing forces did, so slowing them down to rest becomes harder.
In the worst case, the front wheels stop their forward motion entirely while the rear wheels are still sliding. The ensuing driver panic usually means that they stop doing whatever they were doing, and brake: the rear wheels will then pull the car entirely around until the front wheels stop the slide. This is, you guessed it, why it turns 180 degrees and not some other number like 270 or such.[1]
In the case of a handbrake turn, the front wheels follow a curve that is so tight that would cause the back wheels, since they take the turn tighter, to hit the curb or worse. Allowing the rear wheels to slide in the forward direction of motion means that they instead "kick out" (keep going in a straight line rather than following the curve of the front wheels), effectively pivoting the car about its front wheels. So the mechanism is not special: it is just that things which have forward momentum need forces to guide them off of it, and the strong forces which keep the backside of the car guided towards the front have been exchanged for weaker forces which slow down the backside of the car without having a preferred direction.