if Vdd was not 9V but 1000V, would it still be enough to put +5V on the gate to make MOSFET conduct... ?
Yes. This is an enhancement mode MOSFET with an silicon dioxide (glass) insulated gate, so +5V or more on the gate terminal with respect to the source terminal should be enough to "turn on" the drain-to-source circuit if the MOSFET is rated to operate with 1000V between drain-terminal and source-terminal.
If you had paid attention to the video, you would have noticed that an IRF 540 MOSFET was used. This particular device has a maximum drain-to-source voltage of 100 V according to
this datasheet. So raising the voltage from 9 V to 1000 V would likely destroy the device. Most of the specs for the IRF 540 specify a gate-to-source voltage of +10 V to turn the device fully on, but the allowable maximum gate-to-source voltage is ±20 V.
More so confusing is the fact that the "ground" side is actually the source of electrons.
"Ground" is NOT the source of electrons. Electrons are everywhere, surrounding the nucleus of atoms and balancing out the positive charge of the nucleus. That is why almost everything has zero net electrical charge in the world about us. You are actually supported by "electron clouds" pushing against each other.
To get electrons to produce a current in a circuit, it is only necessary to get them moving from one place to another place along a conductor such as a copper wire. "Ground" is totally irrelevant to the process. "Ground," in terms of circuit analysis, is simply a convenient zero-potential reference point against which to measure circuit voltages. By definition, all potentials between "ground" symbols are zero potentials. If you do measure a potential between two different points labeled as "grounds" in an equipment, there is something creating that potential, and neither "ground" should be considered a "true" ground. Even the Earth itself can exhibit differences in potential between two copper rods driven into the ground.
Consider the common dry-cell connected to an LED with a series current-limiting resistor. The dry-cell produces a charge separation at its two terminals by means of chemical action within the cell. One terminal has a negative charge with respect to the other positive terminal. The converse is also true: one terminal has a positive charge with respect to the other negative terminal. The dry-cell, considered as a whole, has zero charge. What it does have is electromotive force (emf) or voltage potential between its two terminals: the ability to move electrons through a circuit externally connected to the dry-cell terminals. The ability to generate an emf and produce an electrical current in a resistive load requires the expenditure of energy, in this case chemical energy produced in the dry-cell. Nowhere is a "ground" ever necessary or required.
Why don't you try purchasing some parts to breadboard a circuit similar to the one shown on the video? Although computer simulation may eventually lead to some understanding, simulation is not reality. You need some "hands on" experience with real parts, making real measurements with real test equipment, such as a digital multimeter, to gain practical knowledge. A multimeter is
essential for electronics experimentation. It doesn't have to be expensive to measure AC and DC voltages and currents as well as resistance to three significant figures. Asian products are cheap and will do the job until you find something with better "bells and whistles."
