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
You can. But only a real battery, not an ideal voltage source. If you short-circuit a battery and check the terminals with a voltmeter, it won't register 1.5V. It will be lower. The same happens to an overloaded power station. It has a maximum power delivery. If the load is greater than that, the voltage will sag (and a real power generator can be damaged).
Transformers aren't ohmic devices. It doesn't have a "resistance" the way a regular resistor does. If you tried to measure the resistance of it in a naive way (with AC current), you would find it has very high resistance if the secondary side could sink a load, and it would have very low resistance if the secondary side did not (for instance if the secondary side were an open circuit).
The behavior of the secondary side affects the behavior of the primary side, which affects the behavior of the primary circuit.
More like "you want the power loss on the wires to be low enough that you can live with it". Given a particular power delivery and a wire resistance, you can calculate what your losses will be on the wire for different voltages.
Your customers demand a particular amount of power. If you don't deliver that much power, the voltage on the line will sag. Given our target delivery, we can divvy that power up with any combination of voltage and current that makes sense. But the higher current will have higher losses.
Let's imagine we need to deliver a megawatt of power to a neighborhood, the total resistance in the line is 5 ohms. Lets see how much power is lost in those lines based on our choice of voltage:
None of these depend on the characteristics of the transformer (other than assuming that we're using it in a reasonably efficient manner).
It is a factor, it doesn't determine it. And it's not an idealized scenario where that is true, it is a different scenario where that holds.
In a simple circuit with ideal voltage and a ohmic resistance, then any given voltage produces a specific amount of power at the resistor. You can do your calculations and they hold.
The big difference here is that the resistor is consuming all the power. In the case of transmission lines we are trying to deliver power elsewhere (the end load). The goal is for the lines to consume less of it and for the end load to consume more of it.
In that scenario, our voltage does not determine the total power, because we will have devices between the wires and the load (the transformers) that change the circuit.
Think of your computer with a universal power supply. Plug it in to a 120V line and it consumes 150W. Plug it in to a 240V line and it consumes 150W. It's not a simple resistance and it's not useful to predict the power it consumes by the voltage supplied.
In a simple circuit where we have excess supply and simple resistance, yes the power consumed is a simple function of the supplied voltage. In those cases, go ahead.
But if you want to talk about why transmission lines benefit from high voltage, that view is not useful. The load is not ohmic and power delivered does not vary with the quadratic of the voltage.