How does the speed of electrons change around a circuit?

If there was nothing in the way then an electron leaving the anode of a 4 volt battery would have a kinetic energy of 4 electronvolts by the time it reached the cathode.

However the mean free path of electrons in metal wires is exceedingly short so electrons never build up anything like this velocity. The end result is that the electron velocities are randomised but with an small average velocity from the anode to the cathode called the drift velocity. To give you a feel for the difference the electron velocity corresponding to 4eV is about a million m/sec but drift velocities are typically around 1 m/sec.

To make an analogy - suppose you are standing in a pleasant breeze of one m/sec. According to the Maxwell-Boltzmann distribution the velocity of air molecules at room temperature is 300 to 400 m/sec but their directions are random and cancel out. You're left with a net motion of the air of 1 m/sec which is analogous to the drift velocity of electrons in a circuit.

The best way to teach electronics to beginners is to use the hydraulic analogy. Even though it might seem clunky at first glance the hydraulic analogy is remarkably useful. You can even model circuit elements like capacitors.

For starters, when we talk about voltage as energy per unit charge, is this energy manifest simply as the kinetic energy of the electron?

Voltage is potential energy per unit charge. An analogy: voltage is to charge as altitude (as on the surface of Earth) is to mass.

So if you lift a 1kg rock off the ground by a height of 1m, you've added some gravitational potential to that rock. Does it manifest as kinetic energy?

Well, if you drop it in a friction-less environment, maybe. But any number of other things could happen. The space within the 1m where the object falls could be filled of some viscous goo. In that case, most of the gravitational potential energy is converted to heat by way of friction. Or, we might design an apparatus to turn a generator as the mass falls, converting its gravitational potential energy into some kind of electrical energy. Maybe we charge a battery, and it becomes chemical energy.

And of course, at the bottom of the fall it probably hits the ground, transferring momentum to the Earth and sending off energy as sound, among other things. It doesn't stay as kinetic energy for long, it most real situations.

And so it is with electricity, too. You probably won't find anything analogous to a frictionless place to "drop" electric charge. If the current goes through a resistor, then it is converted to heat. If it goes through a motor, then it's converted to mechanical energy.

There are devices that launch electrons in essentially lossless conditions (vacuum tubes, CRTs), but most such devices have something at the end into which the electrons smack to do something else (in the case of the CRT, produce visible light). In these cases, the kinetic energy of the electrons immediately prior to impact would be proportional to the voltage through which they have fallen, but I don't suspect this is especially insightful for teaching electronics generally. If anything, it demonstrates that even if the current is constant in a series circuit, the speed of the electrons is not necessarily so.

To know what becomes of the electrical potential energy, you need to know what happened to the charge as it fell.

Do not equate potential with kinetic energy. How fast electrons flow in a conductor has very little to do with their potential. You need to consider the current and the charge carrier density for that. Depending on the material you can have a few fast electrons or many more slower ones. In semiconductors the carrier velocity will be higher - which is why the Hall effect shows up more, for example.

Electrons (charge carriers) lose energy as they are scattered by the material they travel through. This is experienced as resistance. How much current flows is a function of voltage and resistance - so yes when you double the voltage and everything else is the same then average drift velocity of charge carriers doubles (same number of carriers traveling twice as fast). But this velocity is surprisingly low for a good conductor - I recommend you try calculating it to persuade yourself.