OK good old OHM's law tells us that the current is inversely proportional to the resistance.
I = E / R
So, with two wires of equal length and all other things also equal (like the path followed or whatever other factors), but with different resistances will carry different currents. Now, how do those currents differ?
The resistance in a conductor is directly proportional to it's length and INVERSELY proportional to it's cross sectional area. Look about 1/4 of the way down the page:
https://en.wikipedia.org/wiki/Electrical_resistance_and_conductanceR = p (l / A)
p is a constant
l is the length of the conductor
A is the cross sectional area of the conductor
A, the cross sectional area is the full area, not just some fraction of it that is near the surface. It does not matter what shape that cross section may be. A round conductor, a square conductor, and a thin but very wide rectangular conductor (like sheet metal or even metal foil), each with the same cross sectional area and the same constant (p) will have the same resistance per unit length and will have the same resistance. So they will all carry the same current - all else being equal, of course. And combining this with Ohm's law, we see that the current is directly proportional to the cross sectional area (A).
A solid, round conductor, like ordinary solid wire, will have far fewer surface charges than a conductor made of thin foil but with the same cross sectional area. But they will carry the same current - all else being equal. If conduction relied on just the surface charges being distributed in a manner from one end of the conductor to the other that produced an E field gradient, then there would be a lot more surface charges on the surface of the foil conductor than on the solid, round one. And those charges would be a lot closer to the electrons inside the conductor so they would exert a lot more force on the interior electrons which would then move faster. And faster moving charges would mean that more charge would pass a given point in a given amount of time - all else being equal again. So the current in the foil conductor would be greater.
BUT, this is NOT the case. The foil conductor has exactly the same current as the round wire conductor. Or as a square one. Or as a triangular one. Or as a star shaped one. Or as one of any other shape. The current is not changed by changes in the amount of surface area the conductor may have, but by the CROSS SECTIONAL AREA.
This seems to argue rather heavily against the idea that it is only the surface gradient that is causing the interior charges (electrons) to move down the conductor.
A further argument and perhaps a better one would be that it will take a finite amount of time, at least 1/c if I am not mistaken, for the initial field to propagate down the wire. So, regardless of weather the force on an internal charge (an electron) is created by only the surface charges or by both surface and interior charges, both the interior and exterior charges (electrons) near the negative battery terminal will start moving BEFORE the ones that are around the half way point in the circuit formed by the wire. Those charges near the negative battery terminal will therefore BUNCH UP. They will become more dense, both on the surface and in the interior. This will create a net negative region both on the surface and in the interior of the conductor. And this net negative region, this net negative charge will exert a force on the charges in front of it.
In the vernacular, it will PUSH those ELECTRONS ahead.
The exact opposite of this also occurs at the same time on the end of the conductor that is connected to the positive battery terminal except in that case a region of net positive charge is created and it will ATTRACT the negatively charged electrons on and IN the wire there. That attraction will then act on the regions of the wire, both surface and interior, in attracting the negative charges ever further down the wire from that positive battery terminal.
Again, in the vernacular, it will PULL those ELECTRONS toward the positive terminal.
In a time (<= l/c), those two effects will meet at the light bulb in the center of the wire loop and it will then have it's maximum current flowing in it so it will light up at full brilliance.
This is not to say that the light bulb may not also have some current flowing in it before that time. But the amount of that current will depend heavily on the exact physical arrangement of the wires between it and the battery terminals and switch.
In this I am assuming that the switch and battery are connected by a length of wire that is negligible.