The surface is only relevant here in that it gives us a natural reason to have an equation of motion with a 4th power of time. But the equations used are not the correct description of Newton's laws (even though they are very commonly used).

When we say that a particle is at rest or moving with a constant velocity, we commonly translate that to dx/dt = v, some constant scalar (or equivalently, dv/dt = 0). But this is almost a shorthand: in reality, we also have to ask that d²x/d²t = 0, and in general that the Nth derivative is 0 for all N. And the function 1/144(t-T)^4 actually has a non-0 4th derivative, so it isn't describing a particle that no forces are acting on. It is instead describing a particle that reaches the summit and then immediately falls away, as you would expect for any point of unstable equilibrium.

What the construction of the surface does is to make it impossible for a particle that didn't start at rest on the apex to ever reach the apex and remain at rest there. Any particle that can reach the apex will do so with an "accelerated" motion (one where some derivative of it's momentum is non-0), and so will immediately fall away.

You can also look at this just by analyzing the requirement for time reversibility: if a particle is nudged so that it arrives at the apex at a time t = T, and was not at the apex at any time t < T, then it can't be at the apex at any time t > T either.

Or, you can see that the stitch is not natural, or even valid, by looking at those derivatives. The branch with t < T has all derivatives of t equal to 0. The other branch has the 4th derivative equal to a constant 1/6. How would the particle acquire this non-0 [derivative of] acceleration without the action of any force?

We can also look at the dome problem by analyzing the force acting on the particles. For a moving particle, whose trajectory is r(t) = (1/144) (t-T)^4 (that is, one that reaches the summit at t=T), the net force acting on it from gravity + surface normal is F(t) = r(t) ^ (1/2) = (1/12) (t-T)². The force at time t=T is 0, but we can see that the force at time t = T+dt is not 0, so the particle can never rest on the apex. Conversely, the net force acting on a particle whose trajectory is r(t) = 0 is F(t) = r^(1/2) = 0, for any t > 0. So a particle starting at rest can never leave.

By the way, we can reproduce a similar system with a much more mundane surface. Say we have a car rolling on a perfectly flat plane with some friction coefficient k, moving along with speed V. The brakes are connected to an oscillator that gradually activates and releases them. The engine is run at a constant rate, constantly producing a forward force Fe = kV. When the oscillator starts activating the brakes, the car will start experiencing an extra force, Fb, that gradually increases from 0 to kV.

The car's motion will continue in fits for as long as the engine and oscillator have fuel. But, the car will frequently reach 0 velocity and 0 acceleration (when Ff + Fb = Fe, the car has 0 acceleration), and then it will start again. This is because the car is experiencing a non-constant force, which means that its acceleration is also a function of time, and temporarily reaching 0 acceleration and 0 velocity doesn't imply that it will stay in this state forever.