What really happens when you transmit energy through air is charge accumulation. Think of a parallel-plate capacitor - electrons accumulate on one side of the plate and holes on the other side. If you draw a black box around the system, it looks like current is flowing through it. But no significant current is actually going through the dielectric, or you will ruin the capacitor.
Electrical engineers model the phenomenon that Veritasium pointed out as capacitive coupling. In a circuit diagram, we would literally just draw an additional capacitor in between the relevant circuit elements.
In DC, this doesn't really matter after a certain settling time because the capacitor has settled to a certain charge. But in AC (or DC right after you flip the switch) it is non-negligible.
Edited to add - to be clear, there IS an electric field in the air - but the current density is negligible unless you've caused dielectric breakdown.
(It happens also with water, if you have the shower with only very hot water and you open the cold one, the output of the shower changes almost instantly but you need like a second to get the mixed water with that is warm.)
What flows outside the wires is the energy. It's very unintuitive but it's true. Feynman has a nice lecture about it. But note that most of the energy flows very close arround the wires, a very small part wanders far away.
There is an exception when you have a radio transmissor with an antena and a reciver. Then the energy flows just through the air (or vacuum). Also when you have a light lamp or a laser.
Actualy every electric circuit emit some radiation as a bad radio tranmisor. But most of the times you can ignore it.
Since electricity and magnetism are really fields per Maxwell equations, the current flow and other electrical things that we attribute to the inside of the wire are really happening outside of the wires as electric fields.
They have a much better explanation than mine certainly...