Attached article provides a quantitative analysis which shows at DC condition Poynting vector has two components, one along the surface of the wire in the direction of power transfer and the other one perpendicular to the wire directed into the wire, representing power loss dissipated inside the wire due to active resistance.
True enough as a statement on its own, albeit a little bit deceptive maybe.
But if you're trying to use this statement to discredit Veritasium's model, then no, you've only described the values of the Poynting vector exactly on the surface of the wire here. You've neglected to mention what's going on inside and outside the wires in that paper.
Inside the wire (s < a): Poynting vectors run perfectly radially inwards (eq 4), "suggesting" the power flow is just from the "true source of power [Poynting vectors outside the wire]" "burrowing" in from the surface of the wire to where they're needed to perform Joule heating/resistive losses. Zero contribution to carrying power from the battery to the load.
On the surface of the wire (s = a): Yes, two components, one along the surface of the wire in the direction of power transfer and the other one perpendicular to the wire directed into the wire. *But*, although you're correct that the component normal to the wire represents resistive power loss within the wire, you're wrong that the component running along the surface of the wire represents power transfer to the load. Because the surface has no "thickness" to it, the integrated total power flow within this 2D surface running along that very same surface (as described
here) is zero. I.e. the surface integral of a field that consists of vectors running only along that very surface is zero.
Outside the wire (s > a): Outside the wire, the same two components are present: a radial component that's generally heading towards resistive heating of the wire further down, and the component running along the wire. Now that we're in a full 3D region rather than the 2D surface of the wire, we can draw a plane separating our battery from our load and compute the power carried outside the wire.
Performing the process described
here, to break down the power carried from the battery to the load into the three different regions in/around the wire:
Inside the wire (s < a): P = 0 (no Z-component)
On the surface of the wire (s = a): P = 0 (intersection of surface with plane is 1D circle not 2D region)
Outside the wire (s > a): P = nonzero, and I can only presume it would work out to I*V, the power dissipated in the load
The one thing I will grant you though is that if you look at fig 6, you can see that the majority of the content of the Poynting vector is quite close to the wire, because both the E and H fields are strongest there. So I wouldn't be surprised if Veritasium's video had the Poynting vectors drawn a little far from the wires to be truly non-deceptive. Not sure though, the full circuit case might be different. But, it's still true that under the strict interpretation of Maxwell's laws/Poynting vectors, the full entirety of the power is carried outside the wire. (According to that paper at least).
It had to be done.
Truly a thing of beauty