In your question, I see 3 different context, where considering gravitational forces :
a) 2 point-like objects
b) 1 point-like object and one extended spherical symmetric object (not too dense)
c) A auto-gravitating extended spherical symmetric object (not too dense)
a) If you take 2 point-like objects, and take the limit $r \rightarrow 0$, in fact, at some value of $r >0$, you create a black hole, because the ratio $\frac{Energy}{Radius}$ cannot excess a constant value $\sim \frac{1}{G}$ (in $c=1$ units). Note that mass is a kind of energy. So you do not have a problem with $r=0$, because you create a black hole before.
b) If you consider a problem of a point-like object and a extended spherical symmetric object like Earth (not too dense), a theorem states that a object at distance $r$ only feels the gravitational force of masses inside the sphere of radius $r$.
That is, for instance, if the point-like object is inside the earth at radius $r < R_{Earth}$, it feels only the gravitational force of masses inside the sphere of radius $r$.
If we suppose a constant density $$\rho = \frac{M_{Earth}}{4/3 \pi R_{Earth}^3}$$, then the force will be
$$ F(r) = \frac {G m M(r)}{r^2} = \frac {G m (\rho ~4/3 \pi r^3)}{r^2}$$
So, you have a linear force : $$F(r) \sim r$$
So, when $r\rightarrow 0$, nothing bad, about gravitation, appears.
(of course, temperature and pression increase very much...)
If the spherical object is very dense, it is an other story, because you have a black hole, and you may have a "singularity": it is thought that something very bad happens to objects reaching the singularity (tidal forces, roasted, etc..). But you are here in the context of general relativity.
c) The last problem is an auto-gravitating extended spherical symmetric object. I will just give this reference Of corse, as usual, if the object is too dense, you need general relativity, black holes, etc...
The terms are used interchangeably only outside of a scientific context, for example, in your kitchen, in the popular press or poor blogs, and even a few bad textbooks.
In a scientific context, you have it almost correct. Heat is the energy that enters or leaves a system on account of a difference in temperature (no work done).
Thermal energy is a component of the internal energy of the system. It is associated with properties that have a quadratic dependency on some parameter. It includes translational kinetic energy ($\frac{1}{2}mv^2$) as you point out, but it also includes rotational energy ($\frac{1}{2} I\omega^2$), and harmonic vibrational potential energy ($\frac{1}{2}kx^2$). Not included are things that do not have a quadratic dependence on energy. The most familiar perhaps is chemical binding energy (including the intermolecular binding energies in liquids and solids) but there can be others. The total of the thermal energy and the other energies is the internal energy.
The ideal gas particle has no internal structure, so it has no rotational energy, no vibration energy, and no chemical energy. So for that special case the thermal energy is equal to the internal energy, equal to the total kinetic energy of all of the particles in the gas.
There's some confusion about all this due in part to the fact that the basic concepts are introduced with respect to the ideal gas, but the distinction that occurs in applying the concepts to a real gas is often not made clear. A further complication is that it is a challenge to introduce the equipartition principle at a pedagogically early stage.
Best Answer
The difference is that the centre of mass is the weighted average of location with respect to mass, whereas the centre of gravity is the weighted average of location with respect to mass times local $g$. If $g$ cannot be assumed constant over the whole of the body (perhaps because the body is very tall), they might (and generally will) have different values.
I don't see an immediate connection with movement though.