In a TEM mode, why do we have a current flowing into the load? In a TEM mode, if the electric field is perpendicular to the direction of propagation, i.e. electron drift is perpendicular to direction of propagation, why do we have a current flowing into the load?
I'm aware that my reasoning is probably flawed.
 A: UPDATE: sorry for always editing the answer, but my reputation is too low to comment.
In the AC regime, at high frequency, the skin depth is very small and we can neglect the current in the conductors. Then:


*

*The TEM mode (with $E \bot \, k$, where $k$ is along the cable) is defined in the dielectric, where there is no charge flow.

*The Poynting vector $S \sim E \times H$ is parallel to $k$, so the energy ``flows'' through the dielectric along the cable.

*This energy can then be dissipated by charge flow in the load, which is not itself a section of waveguide, and does not support the TEM mode. There is no contradiction here.


In DC mode one needs to assume that the conductors of the cable are perfect, $\sigma = \infty$ (otherwise, $E$ would not be perpendicular to $k$) and then the current flows along them without a voltage drop (which only occurs across the load, hence the dissipation). Even in this case, the Poynting vector treatment is still valid.
I think that in DC mode and in the intermediate regime (AC, but low frequency) an imperfect conductor waveguide cannot sustain a true TEM mode, since the $E$ field lines are deformed close to the metal due to the current.
For more details, see sections 8.1-8.5 in J. D. Jackson, Classical Electrodynamics, 3rd edition (1999).
A: Good question!
Since you are using the term TEM mode, I suppose you are familiar with transmission line theory already. If not, have a look here. For more details consult any introductory electromagnetic books. In my answer I will assume you are familiar with it
As a simple example to answer your question, I will focus at coaxial cable (everything else have the same concept). The next figure shows a transmission line circuit

Assume that it is a coaxial cable (inner conductor surrounded by outer conductor). A cross section of the transmission line looks like this

I suppose everything is clear in the picture, now your question "How come the current is perpendicular to electric field". As you know from the boundary conditions of perfect electric conductor (as assumed here), the normal electric field at certain point gives rise to surface charge density. So in the cross section figure there is a positive surface charge on the surface of the inner conductor and a negative surface charge on the surface of the outer conductor. I hope it is clear up to here.
Now from EM wave theory, we know that the electric field varies in time and in the direction of propagation of the wave, which means that the surface charge density varies  in time and along the direction of propagation of EM wave. From charge continuity equation (equation in the second page of this presentation), we know that a time varying charge density gives a rise to net current (non-zero divergence of current density). In the particular case in the cross section plot, the charge density only resides on the surface so it is a surface charge density. Accordingly, the current should be flowing only at the surface of the conductors not within it. I took this picture from internet so it isn't my mistake.
At the load, the inner conductor is connected to the outer conductor through the load, since we said there is a positive surface charge on the inner conductor and negative surface charge on the outer, they move toward each other through the load. That is the source of current in the load. The way the load is connected to coaxial cable is shown below

Don't forget that the electric field is changing, so the sign and the magnitude of the surface charges on both conductors keep changing.
I hope that helped
