Nothing is wrong with using the work energy theorem for rigid bodies in the case of friction, but as always you need to be careful. The work energy theorem is a bit tricky.
First, you need to distinguish between “net work” and the thermodynamic work.
Thermodynamic work is a transfer of energy by any means other than heat. This is the work that you are interested in when you are looking at the conservation of energy and seeing where energy flows from and to. The thermodynamic work done on an object in Newtonian mechanics is given by $\vec F \cdot \vec d$ where $\vec F$ is the force on the object and $\vec d$ is the displacement of the object’s material at the point of application of the force. The sum of the thermodynamic work for each force acting on an object is the total thermodynamic work.
The “net work” is defined very similarly as $\vec F_{net} \cdot \vec d_{CoM}$ where $\vec F_{net}$ is the net force acting on the object and $\vec d_{CoM}$ is the displacement of the center of mass of the object. The “net work” is only useful for tracking changes in kinetic energy (KE), but provides no information about where that energy came from nor about the total thermodynamic work done on the object.
With these definitions in mind, the work energy theorem states that the change in the KE is equal to the net work: $\Delta KE = \vec F_{net}\cdot \vec d_{CoM}$. This expression holds in general, including in cases of friction. So the work energy theorem is valid, even with friction. However, it does not tell you anything about the flow of energy between objects.
Now, the thermodynamic work is more interesting. Suppose that we have a stationary table and we are sliding a rough block across the table. The block begins with some initial velocity in the positive direction and comes to a stop after a displacement $\vec d$ due to a frictional force $-\vec F$ where the - indicates that it is pointing in the negative direction. Now, the thermodynamic work done on the block is $-\vec F \cdot \vec d= -Fd$. By Newton’s 3rd law the force on the table is $\vec F$ and so the thermodynamic work on the table is $\vec F \cdot \vec 0=0$.
This is interesting. Mechanical work $Fd$ leaves the block but no mechanical work enters the table. The difference is mechanical energy that is lost and is converted into thermal energy at the interface. From that point it is a question of thermodynamics about where the thermal energy spreads.
The contact weld idea is a good “microscopic” explanation, and if you knew the location and displacement of each contact weld then you could replicate that information. However, in practice you never have that level of information, so it is better to simply use the macroscopic information you do have to analyze the energy flow.
In short, “net work” can be used with friction, but it doesn’t tell you as much as the thermodynamic work does. In either case you need to be careful to use the correct forces and displacements for the quantity you want to calculate.
Now, your quote mentions the “particle form of the work energy theorem”, implying that they may have a different expression for non-particle objects. I don’t know what that may be. The work energy theorem I refer to is $\Delta KE=\vec F_{net} \cdot \vec d_{CoM}$. Your textbook may have multiple variants.
Edit (responding to question edit - please do not edit questions in ways that invalidate existing answers):
Notice that as I said above $d$ is "the displacement of the object’s material at the point of application of the force" (emphasis added). Although the point of contact is not moving, that is irrelevant, the material of the belt is moving at the point of contact so $d$ is non-zero for the belt and negative thermodynamic work is done on the belt. This results in the observed loss of mechanical energy of the belt and its conversion to heat at the contact point.
