You need to be careful with analogies. Here are some problems in the analogy you describe:
water starts flowing from a collector (battery)
Nothing in an electric circuit really works like a collector of water. In your analogy, water is electric charge which, in metals, is carried by electrons slowly drifting. Batteries do not store charge, they are not a reservoir of charge (nor of electrons).
Batteries store energy in chemical form. A better analogy is that a primary battery is a coal-fired water pump that will deplete it's store of coal as it pumps water. A secondary battery is a bit like a pump powered by a wind-up spring, it can be run in reverse to wind up the spring. These pumps can only pump water if their outlets are connected to a circuit of pipes that eventually returns to their inlets.
does that mean that the voltage and current increase ...
Voltage isn't something you measure at one point, it's something you measure between two points - it's a difference.
If you measure the voltage at every millimeter of the circuit with respect to the batteries negative terminal you will see the voltage monotonically decreasing as you progress around the circuit‡.
The current measured at any point in the circuit‡ is the same. It neither increases nor decreases
... but the number of electrons would decrease
It isn't very useful to think of the number of electrons increasing or decreasing. Where would they go? Where would they appear from?
You measure a current† of water in litres per second. You measure a current of electricity in coulombs per second (amperes). In a steady-state system, this current is the same in all parts of a simple serial circuit - whether of pipes or of water. A constriction in a pipe cannot make n litres per second of water disappear.
If you slightly turn a gate-valve in a water pipe, the flow of water (litres per second) decreases in all parts of the circuit, including in the pump.
resistance increases causing the pressure and speed of the water to increase but the volume would decrease.
That's not how water works!
If we imagine a simple circuit where a water pump is pumping water around a loop of pipe. The pipe is of uniform size apart from one place where we have a section of narrower pipe.
resistance
The resistance is greater in the narrower pipe (a greater proportion of the water is close to the pipe walls and experiencing friction)
pressure
However the pressure is lower, not higher!
speed
It's the lower pressure that causes the water to accellerate to a higher velocity as it enters the narrow section.
volume
When you say the volume increases, I think you mean the velocity increases. Water is relatively incompressible, it's volume doesn't change much at the pressures applying in our analogy.
The flow rate (volume per second) is unchanged.
Footnotes
† This is one of the areas where the analogy starts to break down. The word "current" is used inconsistently. If you asked someone to measure the current in a river they might give you an answer in metres per second ("current" = average drift velocity of H2O molecules) instead of litres per second ("flow" = litres per second passing a fixed point).
‡ This answer applies only to a simple circuit of battery and resistor connected by copper wires.
I know that watts are equal to amps times volts,
Instead, let's say that Power is equal to the product of current and voltage. It's beneficial to keep the terms seperate from their units.
and that watts are a rate of energy consumtion.
Instead we'll say that Power, measured in watts describes a rate of energy consumption. By the way, this is known as Watt's Law, P=IE.
I have looked at some circuits, but I am unsure of how the total energy use is being distributed. My guess—correct me if I’m wrong—is that the energy used by a component is the voltage across the component times the amps through it. Since the volts add up to the power source voltage, this maintains the total amount of watts.
Yes, this is correct. Note that there are different methods of measuring and describing voltage, current and power, so if you want to actually figure things out, you need to differentiate between, for example, Vrms(root mean square voltage, a form of average) and Vavg(A more literal average voltage), and if you want to work on circuits containing capacitance or inductance, you have to differentiate between P(Real power, measured in Watts, a measure of actual power consumption), S(Apparent power, measured in var(Volt-Amps reactive), the power you will read if you actually measure the circuit) and Q(Reactive power, measured in VA(Volt Amps), which describes the portion of the apparent power that is not actually being used (something capable of storing power is feeding it back into the circuit)
No amps through something, no energy being moved. No volts across something, no pressure difference is necessary to traverse it.
Or, put another way, given that P=EI, due to the multiplication property of zero, if E or I equals 0, so too must P.
Does this mean that a resistor is not just slowing the flow, but turning the energy into heat?
Yes. Note that same as the different types of power mentioned above, for resistance, R is measured in ohms, but if you have capacitance or inductance in the circuit, you have to differentiate between different "types" of resistance. So we add the terms Z for impedance, measured in ohms, and X for inductive or capacitive reactance, also measured in ohms. "Resistance" is used to describe the portion of the impedance that dissipates(uses) power, and "Reactance is used to describe the portion of the impedance that is caused by power being stored and returned to the circuit.
An inductor that is “charged up” and has little resistance is not using any energy because there is no voltage across it, even though there are many amps?
Ermmm... This part seems a bit sketchy. An inductor resists changes in the current flowing in the circuit. If you try to increase current, it will store energy in it's magnetic field, decreasing the rate at which the current increases. If you try to decrease current, it will release the energy stored in the magnetic field to maintain the current flow. Inductors do have (ideally) low resistance, but can have high impedance. The actual resistance and the actual current flow will determine how much power an inductor wastes, whereas the voltage and impedance of the inductor will determine how much current flows in the first place. Note that constant DC power will not experience impedance from an inductor, only it's resistance.
Superconductors are the only conductor available to humankind that does not have resistance. In electrical theory, components are often treated as being "ideal", meaning they have no resistance at all, but in real life, they have as low a resistance as is practical. In order to have power dissipation, either voltage or current must be nonzero, so if there was as you describe, a large current flowing but no voltage drop, you would be describing a superconductor. You're probably wondering about a large current and a tiny voltage drop or a tiny resistance, perhaps too tiny to measure, and yes, if a device has a high current and a tiny voltage drop, that would indicate low resistance, low impedance and low(comparatively) power dissipation.
The vast energy loss in a short is in the existing resistance, the battery/power source?
I'm not sure what you're asking here. A "short circuit" is by definition a circuit with low impedance. Low impedance means a high current, which, without protection mechanisms to break the circuit or limit current, will produce tremendous heat and safety hazards.
When a large current is permitted to flow, it will dissipate power in everything it flows through, proportional to the resistance of each part.
I always imagined (but knew better) that less resistance should be easier. Perhaps it is; the wire is hardly drawing anything. But is the resistance in the source?
Hmm I think you need to clarify what you're asking here, "less resistance should be easier"? Easier for who, the user, the installer, the owner? You could also be referring to the fact that less resistance means electricity will flow more "easily", but if that's the case, "but knew better" should not be there.
Every conductor other than superconductors has resistance. So your electrical source, the wires that connect it to the load and the load itself, has resistance, and they all contribute their resistance to the circuit. So if you have a battery, 2 wires, and an LED forming a circuit, your total resistance would be the internal resistance of the battery, plus the resistance of the two wires, plus the effective resistance of the LED.
One last thing, maybe a bit off topic: is it easier to turn a generator connected to an open circuit than a closed one simply because no energy can be transferred?
Yes. The less electrical load is placed on the generator, the less countertorque it will place on the energy source, so if you're driving it with a gas motor, less gas will be used to maintain the rotor speed.
A water pump would get harder in that scenario. (In terms of work, less energy would be used if I am always applying the same pressure to the crank. It wouldn’t move, so no work.)
The only thing about this I would note is that an absence of work output in a system does not indicate an absence of energy use.
Best Answer
The water analogy is full of leaks. A better analogy, if you need one, is a closed circuit water pumping system such as a central heating hot-water system. Pump = battery, pressure = voltage, radiators = resistors, current = current.
Forget all that for a moment: why does voltage affect power? Increasing the voltage allows us to get the same current through a higher resistance. Since more work is required to do this then the power must be increasing.