Why is the velocity on the top of a wheel twice the velocity of its axle?

I'll tackle your questions in reverse:

3. The contact point is stationary because the wheel is not slipping. This happens when the force of static friction is able to counter the force of the wheel on the ground. This is what you want for controllable transport. If the wheel starts slipping (because of low friction) that's a skid and you are no longer able to steer or brake. If you like, imagine getting your car stuck in mud. You spin the wheel and fling mud all over the place without getting anywhere - not enough friction. If that doesn't help try taking a wheel, marking a spot on it, and slowly rolling it while carefully watching the point of contact.

2. You need sufficient static friction to enforce the no-slip condition. The relation between the velocity at the top, centre and bottom of the wheel is geometrical and is not affected by friction per-se. If the car wheel spins with angular frequency $\omega$, has a radius $R$ and velocity at the axle of $v$ then the velocity of the wheel at the top and bottom is $$ v_{top} = v + \omega R, v_{bottom} = v - \omega R $$

1. If the no slip condition holds then $v_{bottom}=0$, so $ v = \omega R $. Using this in the top equation gives $ v_{top} = v + v = 2v $, independent of $R$. This is because both $v$ and $v_{top}$ increase in the same proportion as $R$ increases.

(Aside: Is there a way to make markdown do reverse lists?)