Voltage is similar to height. It plays the same role for electric charge as height*gravity does for a ball on a hill. So high voltage means high potential energy the same way a ball being high up on a hill means high potential energy.
Voltage is not potential energy, the same way height is not energy. However, if you have a certain amount of charge $q$, you can multiply it to the voltage to get the potential energy, which his $Vq$. This is similar to the way you can multiply height to mass*gravity to get $mgh$ for the potential energy of a ball on the hill. So voltage is potential energy per unit charge the same way height*gravity is potential energy per unit mass.
Voltage must be measured between two points for the same reason height must be. When someone says "the height here is 1000 feet", they are actually comparing it to a point at sea level. In electronics, "sea level" often gets replaced with "ground". So if someone says, "this fence is electrified at 10,000 Volts", they mean there is a 10,000 Volt difference between the fence and the ground, the same way they mean that there is a 1,000 foot drop between the current elevation and the ocean. However, you can use any two points to measure height differences. If you drop a ball, it makes more sense to talk about height above the floor of the room you're in than to talk about sea level. Similarly, if you want to look at a single resistor, it makes the most sense just to talk about the voltage change across that resistor.
The work done on a charge as it moves from point to point is the quantity of charge times the voltage difference. This is just like the work done on a ball as it slides down a hill is the mass of the ball times the height of the hill times gravity.
A single battery cell can only produce a couple of volts. That's how much the potential changes for a single electron in the chemical reaction in the cell. This is a bit like the way a pump that works via suction can only lift water about 30 feet into the air, since that's the potential energy from buoyancy from the entire atmosphere. You can stack multiply batteries on top each other to get a higher total voltage drop (as is done in 9V or 12V batteries) the same way that you could use multiple pumps to suck water higher than 30 feet.
If you increase the voltage across a circuit element, in general the behavior might be quite complicated. This is like saying that if you tilt a ramp to a steeper angle, you will change the way that objects slide down the ramp. In many materials, we find that the behavior simple: current = voltage/resistance. So if you double the voltage, you double the current. This is called Ohm's Law. An accurate description of why it is true is probably a bit too advanced for right now. You will do okay for intuition if you start thinking of electrical current as being like water flowing through a tube. Then Ohm's Law says that if you're powering the flow by having the water flow downhill, if you make the downhill flow twice as steep, the water flows twice as fast. Yes, you can think of it as saying that the electrons are going faster.
Adding resistors in series is like adding several pipes to go through. If you try to push the water through more pipes, it will become more difficult. If you were letting water flow down a hill through a series of pipes, the more pipes you have, the less each pipe can be pointed downhill. That means that adding more pipes makes the water flow more slowly everywhere. Similarly, adding more resistors in series reduces the current everywhere.
The quantity you actually measure when it comes to current is the total flow - number of electrons per second passing through. If you have a 1-ohm, 5-ohm, 1-ohm resistor series, they will all have the same current going through them. This is because if they did not the current would start building up somewhere, and that would change the flow. (This actually happens, just very quickly because the wires have very low capacitance.) The way they all get the same current is they have different voltages. Most of the voltage drop for the entire circuit will be across the 5-Ohm resistor. This is like setting up pipes so that a skinny pipe goes down a steep portion of a hill while two fat pipes go down shallow portions of the hill. The total water going through each pipe per second would be the same. In this case, the water would move faster through the skinny pipe (the high-resistance portion). This is just because the total flow is the same, so if the cross-sectional area is less, the velocity is higher to compensate. This sort of picture roughly works with electrons as well. It is called the Drude model. It is the easiest to visualize, but it is not true to the quantum picture of modern physics.
Batteries do die slowly, yes. That is why flashlights, for example, grow dimmer and dimmer before turning off entirely.
To say a circuit component has a voltage is just saying that there is a certain voltage drop across that element. It is like saying that each pipe in a series of pipes running down a hill has a certain height difference, and that the height difference for the entire system of pipes is the sum of all the height differences of the individual pipes.
If two resistors are in parallel, they have the same voltage drop. This is like saying that two pipes side by side have the same height difference. The one with 1-Ohm resistance will have five times as much current going through as the one with 5-Ohm resistance.
What you could say is "if energy is lost in a resistor, then why doesn't the velocity of the charged particles increase, as per Work-Energy theorem?". So your question should really go something like "Why doesnt current which is $Q/t$ increase if the velocity will increase after voltage drops"?
The answer to this is that we assume all potential energy lost, is again lost due to inner collisions with other atoms, and that's why materials heat up! Also, this implies a steady current because the drift velocity will not be changing.
Edit:
It can be shown that $F\Delta x= \Delta KE$ this means that a force acting on some distance will produce a change in kinetic energy. You can imagine an electric field where the work done by it is simply $W= qE\Delta x$. By the way notice that if I were to divide that expression by the charge $q$ I would obtain the voltage across the thing i'm concerned with. Namely $V=E\Delta x$.
So in summary, if I do some work, then I have some change in kinetic energy. If there is a voltage drop, it follows that positive work was done and kinetic energy increased, which means a velocity increment.
Across a resistor there is a voltage drop. So why isn't it that charged particles are going faster and then I can measure a current increase? Well that's due to the explanation I gave above my edit portion. Namely, that all that increase in kinetic energy is absorbed due to collisions with the neighboring atoms.
By the way, if you're curious enough to visit this website, I suggest you look up a video on Work-Energy theorem on youtube. This concept is pretty straightforward and I'm sure you can understand it.
Best Answer
I admire your determination to understand, and your line of questioning. I'll try and address one or two of your points.
First of all potential difference in free space in an electric field. I prefer to define pd between two points, P and Q, as the work done by the electric field on a charge, per unit charge, as it goes from P to Q. So a greater pd between the points implies more work done on the charge as it goes from P to Q, which can only mean a greater electric field strength, that is a greater force acting on the charge. [Work = Force x distance in direction of force, and we're considering the fixed distance between P and Q.]
So if you apply a pd between the ends of a wire, the free electrons in the wire experience forces, urging them to travel through the wire. Because of collisions between the electrons and the lattice of ions (this is simplified) the electrons don't accelerate continuously under the force from the electric field, but reach a steady mean speed (called the drift speed). If you increase the pd you increase the force on each electron and the drift speed increases. This means that more electrons pass through any cross-section of the wire per second, that is the current increases.
Jumping now to the end of your question: "Why does POTENTIAL energy per charge between two points translate to THERMAL energy? I thought that if voltage was a push, the potential energy would just get converted to kinetic energy in the electrons, so where did the thermal energy come from?"
(1) Voltage isn't "a push"; its units are joules per coulomb! But, as I tried to explain above, it is related to the push (that is the force) that charges get in an electric field.
(2) The thermal energy comes from the collisions that the electrons, driven by the electric field and losing electrical potential energy, make with the lattice of ions. This increases the random vibration energy of the ions. [The extra kinetic energy that the electrons acquire due to the voltage applied is pretty negligible. For a current of a few ampère in an ordinary wire, the drift speed is in the order of a millimetre per second.]