As I've understood it, work is only done on an object if the object
experiences a change in its mechanical energy.
Per the work-energy theorem, net work is only done on an object if the object experiences a change in its kinetic energy. Mechanical energy consists of kinetic plus potential energy. An object does not possess potential energy because potential energy is a system property, not a property of an object.
This means that if energy is added to an object or if energy has left
an object, some force must have acted on the object and thus done work
on it.
Again, this only applies to the kinetic energy of an object and work done is the net work done.
So now onto the question: Let's pretend that we have an object of mass
10 kg and we drop it from a height of 2 meters. Using the formula for
gravitational potential energy (EP = mgh), we get that the object has
a potential energy of 196,4 J before being dropped.
It is the combination of the object and earth, i.e., the object-earth system, that has the potential energy of 196.4 J, not the object alone.
If we choose the object as the system, then the force of gravity is
external to the system. It does external work on the object and thus
increases its energy (kinetic energy).
That's correct. But it is also correct for the earth-object system where gravity is an internal force of the system. For the earth object system, if there is no net external force acting on the system and no internal dissipative forces (e.g., friction) acting on the system, the increase in kinetic energy of the object equals the decrease in gravitational potential energy of the earth-object system.
From the thread I mentioned earlier, it is stated that the object
cannot have an initial potential energy, simply because potential
energy is measured relative to a reference point.
Though I haven't read the thread, again gravitational potential energy is the property of the earth-object system. It depends on the relative position of the object and the earth. It is independent of any reference point outside the earth-object system.
If the object is the system, then there is no reference point. This
makes sense to me, but what doesn't is the fact that kinetic energy is
also relative.
The reference point is the location of the observer. The observer can be anywhere inside or outside the system. So to say there can be no reference point if the object is the system is incorrect.
The difference is gravitational potential energy, since it depends only on the relative position of the earth and object in the earth-object system, is independent of any reference point outside the earth-object system. On the other hand, velocity (and thus kinetic energy) depends on reference points both within and outside of the earth-object system.
The kinetic energy of an object is proportional to the velocity of the
object,..
Kinetic energy is proportional to the velocity squared.
and we all know that velocity is measured relative to a reference point, right?
Correct.
My question is therefore; How can one measure the kinetic energy gained by an object that is falling from a height,
if the object is the system and nothing else is included in the
system?
Again, you can measure the kinetic energy with respect to any reference point. It is purely arbitrary. What is measured will depend on the relative motion between the object and the observer.
Hope this helps.