I have written a detailed answer regarding how a BJT works as an amplifer and this will help you understand how a MOSFET (Metal-Oxide-Semiconductor field effect transistor) works. The way NMOS (N-Type MOS) works is very similar to how a NPN Bipolar Junction Transistor (BJT) works. I wont include much of mathematics here, but it will help you to have a physical intuition.
I am taking NPN transistor for my analysis throughout the answer. Before we go into analysis, we know that a standalone N-type semiconductor and standalone P-type semiconductor acts as a conductor as they have free to move charge carriers, electron in N-type and hole in P-type (well P-type has “moving” holes. A forward moving hole can be treated as an electron moving backward, hence having mobility). Basically, the transistor works because we are forcing the sandwich of N-type and P-type wafers to act as a single “uniform” semiconductor, like when a NPN (NMOS) transistor is in ON state, it acts as a single N-type semiconductor, hence conducting electricity (same argument goes for PNP (PMOS) transistor where it acts as a single P-type conductor).
Let's see how a NMOS works. I have simplified most things and avoided as much mathematics as I could do.
Continuing from the quote from my earlier answer, Here is the simplified structure of an NMOS. It is similar to that of a NPN transistor but with a metel-oxide interface as shown. In the images, S is Source, G is Gate, D is drain and S' is "base" terminal.
We see that the arrangement is kind of a diode connected back-to-back. So now we see that due to the P-type substrate sharing itself with both drain and source it forms P-N junction on both the sides (red) I have added a diode analogy to the right of the MOSFET figure. The P type is lightly doped. Now have a look at the diode picture, in this way, suppose if we apply a potential difference across source and drain or drain or source, the current won't flow because one diode would always be reverse biased. Recall that a standalone N-type semiconductor is actually a conductor, so how can we make to this N-P-N arrangement as a single N-type semiconductor? If we push electrons into P-type semiconductor, it actually behaves as a N-type semiconductor (Why? because the behavior of the semiconductor type actually depends on the amount of majority charge carriers in it). If one is familiar with a bit of semiconductor physics, you can see this through the Fermi-Dirac derivations. So how to make this N - P - N arrangement behave as a standalone N - type conductor ?
How A Metal-Oxide-Semiconductor (MOS) Capacitor works
Consider a MOS capacitor as shown:-
This is basically a capacitor, the metal and P type substrate acts as conductors seperated by an insulator which in this case is a Silicon Oxide (SiO2). I am not going into details of flat band energy and such but lets see what happens if you apply a potential across G and S. This arrangement has a fairly electrically independent capacitance C_mos which is charge inside it divided by potential applied. We know:-
$$C_{mos} = \frac {Q}{V_{gs}} \implies Q = C_{mos}{V_{gs}}$$
Which means the amount of charge pushed into the plates of capacitor is directly proportional to the Voltage applied across the plates due to the capacitor action by the electric field in inside the substrate. Here the plates are the P Substrate and the Metal. When a potential difference is applied across G and S (with appropriate sign), we pump charge (electrons) into the P substrate which increases the electron concentration inside the P substrate. The behaviour of conductor is dictated majorly by amount and type of charge carriers present in it. Since we pump electrons inside the P substate, all holes are filled by electrons and in fact excess electrons start to accumulate inside the P substrate changing to behavior from P type to N type conductor, more charge you pump, the more N it becomes. This process is called the Inversion. Refer the image below, the change in the type is referred as N' 
Lets apply this to the MOSFET structure we saw earlier, when a potential difference is applied between G and S' (In almost all mosfets, the S' is shorted to S which reduces number of terminals and makes S as one of the reference for volage, but introduces something called the body diode which you might be knowing), we pump electrons into the P - substrate. We know depletion region is a result of junction formed between P and N semiconductors, since we are changing the behavior of P substrate to N, the excess electrons nulls the holes present in the substrate, which in turn diminishes the depletion layer, slowly making the N - P - N arrangement as a single N conductor. But if you apply the potential between G and S such that there isnt enough charge carriers inside the P - substrate, the diode action persists leaving the MOSFET non conductive. So, we can say that there is a minimum potential difference that needs to be applied across G and S such that the amount of charge pumped inside the substrate is well enough to make substrate conductive by nullifying the depletion region formed otherwise. This voltage is what we call as Threshold Voltage V_th. I am adding images which shows this:- (The slight red signifies the charge pumped inside slowly diminishing the effect of depletion layer). Recall that the substrate is lightly doped, meaning it doesnt take a lot of charge needed to invert it, thus the required Vgs is also considerably small to make it conductive.
This is how an NMOS works. same analogy you can apply for PMOS as well.
I am not sure why your professor has given that graph to explain you how a MOSFET works. The graph shows the voltage drop occurs inside the P substrate after inversion, which basically acts as a resistor (which we call as Rds). The potential difference across the channel is uniform only if no current is flowing between S and D. Since the S' in my first image is shorted to S, we can say the potential at S' is equal to S, the potential across the channel is not uniform due to the resistor behavior of channel. The resistance and the voltage drop at distance l from Source is given as where L is total channel length (basically voltage divider):- $$R = \frac {\rho L}{A} \implies V = \frac {V_{DS} l}{L}$$.
Hence you get a graph like this. Its worth noting though. As the potential between Vgs varies linearly you have to be carful enough to no make the electric field between S and D below the Voltage required to sustain conduction of current. Suppose the potential drop is too much such that at any point in the channel, the electric field would be too weak to hold the charge in the substrate (or channel) as $$Q \propto V_{gs} \implies Q \propto L - l$$ Therefore, we are making the channel prone to depletion due to the potential drop and loses its effective conductance (comes out of saturation, this is what you see in a Id vs Vds graph of a MOSFET. For a given current Id, there must be enough electron concentration in the conductor to conduct electron smoothly, or else voltage starts to build up, making the MOSFET slowly drift into attenuation / linear region, similar to how a BJT works). That means you need more V_gs to make the electric field at that point enough to pump more charge, sustaining hasleless conduction.
All images are made using Google Draw
