I still don't quite get his explaination at 7:35
He shows the Poynting vector S coming out from the battery in a DC circuit. How? There is no EM radiation loss.
i.e. visually, how does the vector coming out of the battery know where the battery is physically?
There is current flowing through the battery + wires, which leads to magnetic flux/field/whatever (H) surrounding the battery and wire, as shown by the blue arrows.
There is also an electric field (E) formed by the potential difference across the two wires, and it takes the form shown by the red arrows (now, let's not get sidetracked by the fact that an AA battery has a case consisting almost entirely of its negative terminal so that E field [and poynting vectors] would actually look very different to what's shown in the picture, it's not actually an important difference for the sake of your question.)
And using the three-fingers rule to find S = E x H, the Poynting vector ends up pointing away from the battery. So the vector "knows" where the battery is by inferring it from the electric and magnetic fields surrounding the battery, even though those are static.
If you were to repeat the same exercise with a resistor dissipating power (positive voltage still and top but current now travelling down the page), then E would be unchanged but H would be reversed, so S would be reversed (-S = E x -H), and now the Poynting vector would be pointing towards the resistor, indicating that power is being dissipated there.