With deriving $\dot{\vec{p}} = \vec{F}$ and the work-energy theorem, we assume that $\frac{dm}{dt}=0$. Why do we assume this, and is it true for elementary particles? Also, can $m$ have any spatial dependence?

Right, we assume mass is constant in these expressions. In principle, if mass is being removed or added from the system, this derivative will have be included and may indeed result in loss or gain of energy of the system.

Massive elementary particles of a given type do have fixed rest mass; however, rest mass may be converted to kinetic energy (and vice versa) in interactions if particle species change, or if particles are created or destroyed. Total energy will still be conserved in these cases. In such cases, we use relativistic kinematics, which allows description of conversion between mass and kinetic energy. You may have seen this in a modern physics class. We'll be sticking with non-relativistic kinematics and assume mass is constant in this classical mechanics class (although I use relativistic kinematics in my daily life as a particle physicist).

Yes, in a finite-size body, there is a continuous mass distribution which can have mass density varying with position. There isn't in general any problem with treating such a body as an infinite sum of infinitesimal point masses. We'll be dealing with finite-size rigid bodies later in the course.