Consider a wire connected to a battery. Now,potential is analogous to the energy of the particles.And potential in a resistor drops because of the friction inside the resistor(considering there is no friction along the wire and outside the resistor).So the friction determines how much energy is lost.So why does the voltage drop to zero when the current passes from the last resistor it encounters along the wire? I mean,if there are two resistors and you have a battery of 6V,why do the resistors ALWAYS drop the voltage from 6V to 0V?Each resistor should cause a voltage drop analogous to the loss of energy due to the friction.I mean that,sometimes I(the current) should become zero(friction in resistors cause more friction than 6V-considering the convention from voltage to work W=q*ΔV) or the voltage will not reduce to zero(friction in resistors cause energy loss less than 6V).
I think the key thing missing in your thinking is that the energy drop across a resistor is not just determined by the properties of the resistor, but also by how much current flows. The cool thing is that no matter what resistors you put in, the current that flows is such that the potential will fall all the way back down. The reason for this is that electromagnetism is a conservative force, and that means that if you go all the way around you have to get back to where you started.
Think of it like a roller coaster that goes up and down hills--if the roller coaster starts at the top of a hill, rolls down without friction, and comes back around, it should get back to the top of the hill without any "extra" speed. Otherwise the rollercoaster could just go around again and again and get more speed for free with no input! In this analogy, resistors don't work like a friction force, they work more like the hills. You can put a charge on top of a hill (a high voltage place) and it runs out of steam (gets to 0 V) right before it is pushed back "up the hill" (i.e. crosses the voltage source)