Electronic – Why do we not use both E and B fields for a receiving antenna

When the EM wave illuminates the resistive sheet, it will induce currents to flow in it. These currents will radiate new EM waves, the sheet becomes a new antenna. As such, it will have both a near field (non-radiating) and a far field (radiating) set of waves travelling outwards from it.

The impedance at the barrier is not 377ohms. The current you'd calculate there would have a different ratio. Only several wavelengths into the far field would the non-radiating (evanescent) wave have died away, leaving the clean radiating wave, which would by now have its 377 ohm electric/magnetic ratio.

It would be a complicated job to compute currents in the barrier given the incident EM wave, especially due to the reflected wave, the partial transmission through the barrier, and the re-radiation from both sides, all with an impedance quite different from that of free space.

This sort of stuff is done with field solvers, rather than equations.

It really depends on the structure of your wavegide at microwave frequencies, you can imagine the fields like the waves in a wave pool, there is a resonant frequency depending on the width and length. But it's more complicated than that because it's a 3 dimensional problem.

The first picture you show applies to all EM waves, because any EM wave is just a collection of photons super imposed together, but when you get to lower frequencies like microwaves the waves are larger with respect to other things and the effects are different (ie a light photon will bounce off of a atom, but a radio wave will not, conversely a radio wave will reflect or be absorbed into a conductor of fractions of a wavelength)

To really visualize the fields you need a 3 dimensional field solver (or be really handy with a calculator with lots of time). The picture below shows a hollow rectangular waveguide, and shows the different modes. There are vertical and horizonal modes and first and second order modes, which in reality are superimposed on top of each other. For design it can be easier to consider one mode at a time.

Source: https://www.cst.com/academia/examples/hollow-rectangular-waveguide

The modes can be calculated on paper and look like this for a 2-d cross section: Source: http://www.wikiwand.com/en/Waveguide_filter

I think your problem is you need to think of wavelength as a characteristic of EM waves

As far as your other question goes if you think of the wave traveling perpendicular to the aperture, the hole edge on would be paper thin, and the wave wouldn't see it (as soon as you add conductors around the hole things get more complicated because waves do change with conductors, but you get the idea)