Why do skidding blocks tip over? If a block sliding on a low-friction surface suddenly hits a high-friction surface, it will topple over (if the friction is high enough). Unless I'm mistaken, the axis of rotation is the block's leading lower edge. In the block's reference frame, this is due to the fictitious inertial force acting at the center of gravity, providing a torque about the leading edge. However, in the inertial frame, the force is friction, acting on the bottom of the block. My understanding is then that there would be no torque, as the force would be directly through the axis. Clearly I'm misunderstanding something, so what is it?
I initially realized my lack of understanding in the context of cars rolling over on sharp turns but thought the block would be simpler because there are only two frames to worry about, with a simpler transformation between them.
 A: 
My understanding is then that there would be no torque, as the force would be directly through the axis. Clearly I'm misunderstanding something, so what is it?

A torque isn't needed to make the block tip over. The block has non-zero angular momentum with respect to where the leading edge of the block suddenly comes to a rest. You are correct that a force applied at that edge exerts zero torque on the block. No torque means angular momentum is conserved (at least briefly), making the block start tipping. Now a new torque arises, which is gravity. If the block wasn't moving fast enough, the block will tip up a bit only to fall back over. If the block was moving fast enough, it will tip. And then maybe tip again. And again. Eventually friction will bring it to a stop.
A: This is more of a dynamics question, but let's see if I can still explain this in a way where only forces are considered :) 
So, initially, the block is sliding unperturbed:

Note that there is no friction ($F_w = 0$), I've omitted the weight ($F_g = m\mathbf{g}$, from center of mass downward), and I've drawn the normal force opposing the weight as a distributed load, which will be helpful later on. As usual, the resultant of the weight and normal force balance each other perfectly, so the block's linear and rotational velocity does not change. 
Then, suddenly, a force $F_w$ is exerted on the leading edge that is in contact with the table. Now, if the block were flying in space and there was just the block and that force, the block would start to accelerate to the left (non-zero total resultant force, Newton's 1st law) as well as start rotating about the center of mass (total resultant force does not pass through the center of mass, causing a torque and resulting change in angular momentum). 
These actions of course also happen while the block is on the table, but, that rotation can not take place about the center of mass, because the table surface is in the way. In other words, the normal force will change such that it will oppose that rotation (Newton's 3rd law). 
Therefore, moments after the friction force is encountered, that induced rotation will appear to increase the block's weight on its right half, while decreasing that of the left half. This is best illustrated as a change in the shape of the distribution of the normal force:

NOTE: in the idealised case it is a simple linear distribution. But the main point is, it is no longer distributed evenly. 
The resultant of this normal force distribution will not go through the center of mass anymore, but it will have shifted forward. At the instant depicted in the drawing, it will also not quite go through the lower-right leading edge -- it will be somewhere in between.  
The resultant normal force will be larger in magnitude than the weight (due to the added opposition to rotation), resulting in a net upward movement of the block. 
It will also cause a net torque about the tipping point, but this is not the main reason the block starts rotating -- that is mostly due to the friction force not passing through the center of mass and changing the angular momentum. The normal force just prevents all parts of the block from passing through the table.
Now, the force distribution will (normally very rapidly) converge to the point where $F_w$ is applied. It will then encounter the following situation:

The block's weight will of course cause a torque about the contact point, opposing the rotation. Depending on the instantaneous orientation, the resultant of the friction force and normal force ($F_R$) may


*

*have a contributing torque about the center of mass

*go exactly through the center of mass (no torque)

*have an opposing torque about the center of mass


What happens next depends on the exact kinematic parameters and details of the setup.
If the block's speed is not large enough and/or the friction is small, the block will be lifted slightly, then drop again and come to a halt.
If $F_W$ can grow without bound (e.g., the block collides with a fixed obstacle) and is moving fast enough, the block will fly up and spin in mid-air. 
If this happens (or some combination of the last two cases) and there is residual forward movement, the whole process may be repeated until the block comes to a halt. 
A: Take a look to the following sketch:

The idea is that the block will tip over when $ΣF*h>W*a$ 
EDIT: This is quite tricky indeed.
Let me provide an inverse point of view situation:
Consider the block at rest and that the ground is starting to accelerate horizontally. What is the acting force  and the acting point on the block?
A: When the block first starts to slide on the high-friction surface, that frictional force starts the block rotating suddenly. If it is enough, the block will rotate far enough such that the center of mass will tip forward and create its own torque around the toe and cause the block to topple forward. Mathematically, it's not a trivial problem and laypeople would be amazed by the calculation.
