Strictly speaking there are no reversible processes in Nature; it is an idealization that enables one to get bounds on efficiency of nonequilibrium processes by using techniques of equilibrium thermodynamics only.
A reversible process is therefore primarily a theoretical concept for discussing what would happen in a process if dissipation were absent. It is defined as a motion in equilibrium state space, and hence presupposes that the system always remains in equilibrium.
However, empirically, fast processes generate far more excess entropy (the source of the dissipation) than slow ones, so one can treat a slow process approximately as a reversible one.
In a nonequilibrium state, the extensive variables are still well-defined, while the intensive variables (temperature, pressure, chemical potential) typically aren't. On the other hand, most nonequilibrium processes in ordinary life are well-described by local equilibrium; which means that every small region is approximately in equilibrium, and then temperature, pressure, and chemical potential can be assigned definite values. As a result one gets a temperature field, a pressure field, etc. (This is what you feel when you move through a room from the heating to a open window.)
This is a very interesting, important and at the same time subtle matter which is useful not only in thermodynamics, but other areas as well, and not everyone quite understands it.
First off, two statements that you'd better just memorize for now:
- In a reversible process, equilibrium is never left. Yes, that is a paradox. That's why all real processes are irreversible. However, a real, irreversible process can approximate a reversible one.
- A reversible process does not waste work: no other process can do more work going from state A to B of a substance than a reversible one. That also mean that all irreversible processes are wasting work, that is, when carrying out an irreversible process, you're losing the opportunity to have even more work done.
Equilibrium, Reversibility and Driving vs Opposing Forces
Let me try to give you an informal but intuitive notion of what reversibility has to do with equilbrium and balance between driving and opposing forces.
Equilibrium is achieved when a driving force in some direction is countered by an opposing force in the opposite direction. The concept of force here can be expanded to include whatever wants to steer any situation to a particular direction. Let's work with the example you gave:
Gas in cylinder with pressure $p_i$ under piston of area $A$ over which lie one thousand lead shots each weighing $w = \frac{p_iA}{1000}$.
The gas wants to expand (driving force), but the thousand shots exactly oppose it (opposing force). All is well and in equilibium here; nothing moves or changes.
Now imagine that you removed one shot. For a brief moment, the force in the piston due to the gas pressure would be marginally bigger than that due to the shots' weight, such that the piston would accelerate upwards a little, thus expanding the gas. This expansion would lead to a small decrease in pressure, which would eventually equal the current weight (99,9% of the original). In other words, for a brief period of time, the driving force trumps the opposing force. Then we have equilibrium again.
If your goal is, say, $p_f = \frac{1}{2}{p_i}$, you can keep doing that until only 500 lead shots are left. Each removal is one step of your process, and note that every step is a small period of disequilibrium between two equilibrium states. We can say that our example process is an approximation of a process that never leaves equilibrium (in this case, mechanical equilibrium!). Also note that our process did some work!
On the other hand, what would happen if, instead of patiently removing one lead shot at a time, we abruptly took away 500 shots? Well, suddenly the driving force (gas pressure on piston) would be twice as big as the opposing force. The piston would crazily accelerate upwards and the gas would expand rapidly. After quite some time, of course the same state would be reached: $p_f = \frac{1}{2} p_i$, $v_f = 2 v_i$. Note however, that the work done is smaller! In this case, only 500 shots were elevated to the final height, while, in the former, the same 500 shots were elevated PLUS 500 others were left along the way, at each height of each step of the process.
The former process (slow) is a better approximation of a reversible process than the latter (fast). Note that, when something is at an equilibrium, a very small push is all you need to tip the state in the direction you want, minimizing time spent in disequilibrium. You don't need to roundhouse-kick it to the desired direction. The more the driving force surpasses the opposing force, the greater the irreversibilities of the process. However, it does happen faster that way.
We can also think of an example involving thermal equilibrium. Suppose you want to raise the temperature of an object from 0 to 100 degrees. If you just put it in contact with an object at 100 degrees, you'll have a driving force disproportionately bigger than the opposing force — that is, nature is DYING to heat that cold object. However, if you used one hundred objects, each at a different temperature, to incrementing our temperature 1 degree at a time, the process would be more gentle. We stay in equilibrium more. Our newfound intuition tells us that this process is closer to a reversible one.
Conclusion
We can sum this up answering your points:
- A step would be a period of time where the system moves toward its goal. In a reversible process, the driving forces (e.g. temperature or pressure differences) would be so small that each step would take forever. In a real process, generally, the "more reversible" it is, the longer each step will take.
- This is an idealization which can be approximated by having a series of "small disequilibriums" instead of a large pressure difference in each side of the piston (akin to a free expansion)
- These are abstract terms (i.e. may not be real forces). The driving force is that which leads the process in the direction you want. The opposing force is that which leads it in the reverse direction. Like pressure differences in a mechanical problem, or temperature differences in a heat transfer problem. The greater the difference between these forces, the farther from equilibrium the system will be, and the process will generally happen faster and more violently. This also means a less reversible process.
Best Answer
For work to be performed, an imbalance would have to be created. That means, forces would be imbalanced and the velocity of something somewhere would have to change. A system in equilibrium won't perform work on anything.
In your case, you could add mass to your original mass and that could create an imbalance and cause the piston to compress. Or, you could remove mass and cause the piston to expand. The act of adding or removing creates an imbalance.
Also in your case, you could heat or cool the gas. That would also create an imbalance causing the piston to compress or expand.
The original system won't do anything unless an "imbalance" is imparted to it. Nothing will happen spontaneously.
The problem you are describing sounds much like the method of "Virtual Displacements" that's used frequently in structural analysis. That method considers a system, already in equilibrium, that is given a virtual displacement. Since the system is already in equilibrium, the forces within the already-stressed system perform no net work whatsoever.