To amuse myself, I had a go at solving this particular problem of overturning the rover.
First, we need to decide how to quantify "how difficult" it is to overturn the rover, whether here or on the Moon.
One way is as described above - will a sharp turn at maximum speed impart enough torque to lift the wheels?
If the centre of gravity is located a distance y (vertical) and x(horizontal) from the inner wheel, and the turning radius is R, the vehicle will lift from the ground if v2/R * y > g * x. This should be simple to evaluate using no more than high-school math. Now, even if you get the outer wheels to lift, it would be very simple for the astronauts to correct - just turn the wheel the other way and the rover rights itself. Someone with more patience than me can calculate for how long you need to go at max speed and turn rate to flip the rover completely given the various scenarios. However, it's easy to see that the necessary speed to overturn is proportional to the square root of the gravity, so that it takes roughly 2.5 times as much speed to flip the rover on the Earth as on the Moon.
The other way has to do with hitting bumps on the surface - rocks etc. It would be difficult to model exactly how the force impacts the rover, but we can simplify by thinking in terms of stunts. One method often used in stunt filming is using ramps, so that a certain horizontal speed will impart a corresponding vertical speed. This means that on both Earth and the Moon, the same initial motion can be used - gravity does not impact this.[1]
To model exactly how the rover behaves here is unfortunately very difficult - the rover can be seen as a pendulum pivoting around the pair of wheels on the ground, but you quickly find yourself having to solve a nonlinear differential equation. However, we can sidestep this completely by simply asking ourselves how much rotational kinetic energy the rover needs to obtain an angle of 45° (at which points the wheels would start to crumple from the sideways load - the actual overturn angle is higher)
This turns out to be very simple. Rotational kinetic energy is simply Er = 1/2 I w2, and since the angular velocity is proportional to the horizontal speed when we hit the ramp, Er = 1/2 I C2 v2, where C depends on the exact geometry of the situation.
To find the moment of inertia, it needs to transposed to the axis we are using from the usual centre-of-gravity oriented one, so using the x and y from before and taking the angle b = atan(y/x), and radius r = sqrt(x2 + y2), Iwheel = ICoG + m * r2
For potential energy, we need the difference between the initial state with both wheels on the ground and the one with the wheels at 45 degrees angle: Ep = m * g * (r sin (b + 45°) - y) [2]
Summing up, the rover will crumple or flip if Er > Ep, or 1/2 I C2 v2 > m * g * (r sin (b + 45°) - y)
Now, there are a lot of factors in all this, but we're interested in comparative scenarios. Everything except g is the same, so again, the necessary velocity is proportional to the square root of the gravity, or 2.5 times more difficult on the Earth as on the Moon.
[1] In reality, the suspension would take up much of the impact, and on the Earth the springs would be compressed more by the gravity load, meaning they're more likely to behave nonlinearly. In other words, on the Earth the same bump would impart a greater force on the rover, but the difference is too small to matter in this equation.
[2] By definition, y = r sin b
