Does energy transfer from the battery through the load through B-fields, i.e. "electromagnetic current" and not really through any other means?
The vast majority of energy transfer is from the electromagnetic fields (E & B-Fields which are coupled). There might be some fringe situations where ion or electron (-e) transfer contribute to total energy transfer, but I can't think of any off the top of my head. That's the primary point of the page you've linked, and the POYNTING diagram in figure 7 should show you the energy transfer (that's what they do). Notice that it shows all arrows pointing to the load and away from the source.
Is "Electricity" simply heavily concentrated electromagnetic fields when flowing through a conductor? For example, an arc between two high voltage electrodes is the electromagnetic energy jumping the gap, the heat and light of the arc are just by-products of this?
Well, this question isn't well phrased, so I'm going to assume that you are asking what is the principal method of energy transfer, which is the same as your first question.
Directly speaking, electricity is the flow of charged particles. The other point to make here is that the electromagnetic fields flow AROUND a conductor, not through it. Someone might nitpick about this, but it's essentially true.
As to the arc, it is caused by the electromagnetic fields exciting atoms of air to the point where you see a reaction. This is akin to asking, "What is fire? Is it just a byproduct of Oxygen combining with Carbon?". Yes, it is, but what you see is an excitation of gas molecules around that combustion.
What if the path were something like a wire surrounded by a magnetic insulator, would the e-field be pointless, or not "care" that anything is blocking it just existing on either side to create a difference in potential?
Would there be less energy available? I am greatly interested in how energy is transferred through the wire, at least correctly.
Here you need to understand that E and B fields are COUPLED. So, no, the E-Field wouldn't not (double negative, sorry) "care", it would be affected and so would the flow of electrons in response. What you've mentioned here would actually be a passive inductive element in the circuit! Look up the theory around inductors and you'll get a good idea of what's happening here.
Does the insulation of wiring (i.e. plastic) also help in reflecting some of the energy so that it does not drop over a distance? This could sense, being that an exposed wire may pick up energy in the air, and an insulated keep it out.
I'm not exactly sure how to respond to that one. The real job of the insulation is to keep the circuit in isolation from the environment, so your second sentence makes some sense, but reflecting energy isn't the appropriate way to think of what it is accomplishing.
Keep in mind that the wire is a metal and thus already has a great deal of free electrons. The battery is not a source of electrons, it's a pump to move what is already there.
Still, you can view the system as a very small capacitor, with the wires being the plates. They would be relatively far apart and have very little surface area so the capacitance would be very very small. So, yes, I think there would be electron transfer, but it would be negligible and it would only last an instant.
Best Answer
Interesting question.
This is one of those questions that gives most EEs a bit of headache because it is rather difficult to get one's brain wrapped around what is theoretically going on with electromagnetic waves. The truth of the matter is it is not quite as simple as we first imagine.
Here is how I rationalise it.
First and foremost you need to separate the notion that electrons have anything directly to do with electromagnetic waves. They don't. EM waves propagate without the need for any material. In a vacuum they propagate at the speed of light, when the wave encounters a material they are slowed by the materials physical properties.
Having grasped that bit of wisdom, it is easy enough to understand that when you apply a voltage to one end of a bit of wire, it takes time for you to be able to detect that voltage at the other end of the wire. That voltage propagates down the wire at the speed of light for that conductor material. In effect you created, or more accurately, changed the E-Field at one end and the change wave takes time to get to the other end.
Now consider instead modulating the applied voltage, that is, applying a signal to the wire. If we break up that signal in the time domain into infinitely small periods you can see from what we just discussed that there will be that propagation delay for the instantaneous change at one end to reach the other. The E-Field must "carry" those changes to the other end. Again, it is important to remember, this has nothing to do with electrons.
So, to summarize so far, the electromagnetic waves are the carrier for your signal not the signal itself. EM waves happen to be transverse, and that does not change.
The signal, or whatever voltage wave you are transmitting, is effectively a modulation of that carrier, and is usually longitudinal. You are setting up, or transmitting, a difference in the electrical field. It's those local differences that excites the electrons in your conductor and make them move in reaction.