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.
Audio inputs are normally AC coupled (there is usually a series capacitor to block any DC component in the input). Typically this means you won't see much below about 20 Hz and you certainly won't be able to measure DC or slowly varying signals.
As for sensitivity, typical "live level" inputs expect a signal of 775 mV RMS which corresponds to 0 dBV. Microphone inputs are usually more sensitive than this, but there is no "standard" sensitivity and the input hardware often has some kind of controllable gain stage prior to the A-D converter.
Best Answer
Since the flow of charge through anything constitutes current, and since current creates a magnetic field, the short answer to your question: "Can an electromagnetic field be created in a non-metallic coil?" is "Yes".