Electrons move because they are in a region of space with a non-zero electric field. They don't accelerate to high speed in a wire because they keep bumping into things; a kind of friction which dissipates energy much like the friction you are used to that explains why resistors get hot. In effect their speed depends on the strength of the local electric field and the nature of the material they move in.
When you hook up a circuit of some kind to a voltage source (battery, generator, wall-wart, whatever), the electric field already present between the terminal of your power supply causes some electrons in the wires to move around. In the course of doing that the electric field gets re-arranged to point along the wires and through the components and so on. There is rather a lot of shuffling that goes on immediately after the power supply is hooked up and I am going to mostly ignore it to focus on what happens when a (short-term) steady-state is established.
Circuit basics
Observe the circuit in its operating state: the electric field points along the wires and though components in various way. In some places that field $E$ is weak and in some places it is strong, and in some places the electrons flow fast and in others slow, but there are two rules that must be obeyed:1
The current (number of electrons passing a point) is the same throughout the circuit. This follows because I have restricted out consideration to time when things aren't changing, and if more were passing point A than point B (a little further down the circuit) electrons would be piling up in the space in between them.
The total voltage change around the circuit has to be zero. This is because voltage is a function and can have only one value at any point in space, so if I follow any path that comes back to itself the changes has to equal zero when it returns.2
These rules are written in terms of voltage and current, but previously I was talking about electric fields, so what's the relationship between them?
The current comes into play in the form of Ohm's Law: $V = I R$.
The potential change in a section of circuit with length $d$ and constant electric field $E$ is $\Delta V = E d$, so we can write the voltage rule as
$ 0 = V_{ps} - \sum V_i = V_{ps} - \sum_i E_i d_i$ where $V_{ps}$ represents the voltage gain of the power supply.3 Rearranges this gives us
$$ V_{ps} = \sum_i V_i = \sum_i E_i d_i \,.$$
One last thing before we're ready to answer the question: the electric field in the wires is usually assumed to be very small compared to the electric field in other things like resistors. Therefore, we can ignore the $Ed$ contributions from the wires in working the maths. This isn't true for very long wires or for very fine wires under low voltages, but we're going with it anyway.
How do the electrons "know"
Consider a very simple circuit with a switch in it. A resistor (numbered 1) is connected directly to the battery and to the input of the switch. From the switch the current gets back to the battery either directly or through a second resistor (numbered 2).
The circuit starts with the switch set so that only one resistor is involved. When we hook it up, the fields rearrange themselves such that we have very weak fields in the wires and a very strong field in the resistor: $V_1 = E_1 d_1 = V_{ps}$. To make the current rule work, we have a lot of electrons moving slowly in the wires and a few electrons moving very quickly in the resistor (think of car flowing).
$t = 0$ The original state of the circuit has a field in resistor 1 $E_1 = V_{ps}/d_1$ and very weak fields everywhere in the wires. There is no build up of electrons anywhere and the current flow is steady throughout.
$t = 0 + \epsilon$ The switch has changed state, but the electric fields have not re-arranged yet, so there is zero field in resistor 2. Electrons continue to move through resistor 1 at the same rate as before, when they get through it there is no field to move them through resistor 2. They begin to pile up between resistors 1 and 2. As they do that they begin to reduce the field in resistor 1 and increase it in resistor 2.
$t = 0 + (2\epsilon)$ Now there is a little field in resistor 2 and a little less in resistor 1. Current has started to flow through resistor 2 but there is still less than is flowing through resistor 1. More charge is building up between them and that is driving the field in 2 up more and the field in one down more.
$t = 0 + (\text{several }\epsilon)$ The field in resistor 2 has risen and the field in resistor 1 has dropped until they are almost matched. Current flow through the 2 resistors is almost the same with only a small amount more coming through resistor 1. The charge between them has almost, but not quite stopped changing and that means the fields in them are also almost fixed.
$t = 0 + (\text{many }\epsilon)$ The field in resistor 2 has risen high enough that it's current matches that in resistor 1. This represents the new current of the circuit as a whole and is lower than the original current.
What we learn from this consideration is that any time the flow of electrons is faster through one part of the circuit than another, electrons will accumulate in such a way as to re-distribute the electric field in the circuit so that the flow is more even than before, and that this process happens continually until the flow becomes uniform throughout the circuit. The strength of the electric field is also related to the voltage change over each component and will be adjusted until the total is equal to the supplied voltage.
1 Rules written in a form that applies only to series circuits. For a more complete version, look up Kirchoff's Laws.
2 This is true when you neglect magnetic induction.
3 I am assuming that there is only one power supply. The full treatment of Kirchoff's laws can relax this restriction.
You may have found a small glitch in that water fall analogy. An analogy I like much better is to think of water through pipes.
The voltage (potential difference) corresponds to the pressure difference between two points. A higher pressure in one spot means a larger "push" on the water. For charges in a circuit, the voltage is the "push" that squeezed them forward through the obstacles in the form of resistors and other circuit components.
Such a pressure difference is directly corresponding to a larger potential energy difference. This is why the water fall analogy is often used, because it is a more intuitive way to think of potential energy. But when you are increasing the voltage across two points in a circuit, then this corresponds to not a higher "pressure" difference from the top to the bottom of the water fall, but rather to a larger potential difference. And such a higher potential difference means a higher water fall, because the potential energy we are comparing with here is gravitational.
So, the increase in height of the water fall is analogous to an increase in charge accumulation in an electric circuit. The distance is changing in that analogy so the speeds are not really comparable. But they would be in the pipe-analogy.
Best Answer
Power to a water-wheel depends both on the current (amount of water delivered) and the head (vertical drop of water as it turns the wheel). So, the water analogy does have TWO variables that multiply together to make power: current, measuring (for instance) the water flow at Niagara, and vertical drop (like the height of Niagara Falls).
Current is NOT the same as power, in a river, because long stretches of moving water in a channel don't dissipate energy as much as a waterfall does. Siting a hydroelectric power plant at Niagara Falls makes sense. In the analogy to electricity, a wire can deliver current at little voltage drop (and has tiny power dissipation) but a resistor which has that same current will be warmed (it has a substantial terminal-to-terminal voltage drop).