The formula you've referenced, $\vec a_c = -\omega^2\vec r$, must define $\vec r$ as a function of the angular position, which depends on time (uniform motion and all). Something like $\vec r(\omega t) = r(\hat x \sin \omega t + \hat y \cos \omega t)$.
$\omega$ essentially represents the angular speed, which means for constant $\omega$ you have an orbital speed $v$ that depends on distance $r$ through the relation $a_c=\frac{v^2}r$. This applies to any uniform circular motion.
The force of gravity at a given distance gives you the acceleration, and with a suitable orbital speed the motion will be uniform. However, as you've noticed, the angular speed for uniform circular motion due to a force $\propto \frac1{r^2}$ (like gravity) will have decreasing angular speed with increasing $r$.
If you're only concerned with the magnitude of $a_c$, this can be shown as follows:
$$a_c = \omega^2r = \frac{v^2}r$$
First, you can observe the following:
$$v^2 = \omega^2r^2$$
$$v = \omega r$$
But you also need to be mindful that $a_c$ itself depends on $r^2$ via $a_c = \frac {\mathbf F_g} m = \frac{GM_e}{r^2}$.
So orbital speed depends on $r$, and angular speed depends on $r$, for uniform orbital motion due to gravity, but in different ways:
$$\sqrt \frac{GM_e}{r} = v$$
$$\sqrt \frac{GM_e}{r^3} = \omega$$
So in a way you're right, if the angular speed is kept constant, you need a larger $a_c$ with increasing $r$ to create uniform orbital (circular) motion. However this is only because the orbital speed increases with distance for constant angular speed $\omega$, which in turn requires a larger $a_c$ in order for the motion to still be uniform and circular. But under gravitation, $\omega$ is not constant because the force doesn't grow linearly with distance (it shrinks in fact), and $a_c$ is lower with increasing $r$ as a result.
Regarding your two probes, let's call them E (on earth) and S (in space)... Consider that E is not actually in uniform circular motion at all - it is on the surface of the earth, which is itself spinning. In fact, it is the other way around, probe E is moving too slow to achieve uniform circular motion around the earth, and would fall down.
Think of it this way - if the acceleration felt is larger than what is required for uniform circular motion at a given distance and speed (as is the case near the earth's surface for every-day situations), the object would approach the center, in other words fall to the ground. Consider how far and hard you have to throw something at ground level into the horizon for it to never hit the ground; geosynchronous / geostationary orbits are exactly this, but at a height where their required speed for uniform motion is small enough to match the speed at which the earth rotates.
It may help to try and work out the orbital speed (not in terms of $\omega$) for S, and then compare it to the orbital speed needed for uniform motion under gravity at smaller and smaller distances to the earth - you should find that as you come closer, you need to be moving considerably faster to maintain uniform circular motion ($v^2 \propto \frac1r$). And with any increase in speed, for the motion to remain uniform and circular at the same radius, the acceleration must increase ($v^2 \propto a_c$).
