The explanation is using an energy argument. That for the normal case of a submerged piece of wood, you can assume that if the wood and a parcel of water above it switch places, then the water (which is heavier/more massive) drops in the gravitational field releasing potential energy. This release is not offset by the rising wood since it is not as massive. This energy imbalance goes into the kinetic energy to move the log (and the water).
This argument says that in the case of the rotating log, none of the water can move downward to release potential energy. Since the water stays in the same place, no energy is released, and there is no energy to move the log.
Why doesn't the object just stay where it is?
Because forces will combine to move it to a position with less energy. You can do a free-body diagram of a ball rolling downhill to see that the net force is down the hill. But you can also simply say that the ball is going to move in a way that lowers its potential energy (which is downward in the gravitational field). Using that argument for the cylinder says that there is no position it is free to move to that lowers the potential energy, so no movement will happen.
Thus, the air molecules contribute a small portion of their kinetic energy to the paddle, which is then expended as heat on the other side of the border, making the air molecules on the left colder, while air molecules on the right heat up. Doesn't this mean a decrease in entropy?
Yes it does.
However, we need to take the thermal noise of the resistor into account.
Hot resistors make noise
As discovered by John B. Johnson in 1928 and theoretically explained by Harry Nyquist, a resistor at temperature $T$ exhibits a non-zero open circuit voltage.
This voltage is stochastic and characterized by a (single sided) spectral density
$$S_V(f) = 4 k_b T R \frac{h f / k_b T}{\exp \left(h f / k_b T \right) - 1} \, . \tag{1}$$
At room temperature we find $k_b T / h = 6 \times 10^{12} \, \text{Hz}$, which is a ridiculously high frequency for electrical systems.
Therefore, for the loop of wire and resistor circuit in the device under consideration, we can roughly assume that
$$\exp(h f / k_b T) \approx 1 + h f /k_b T$$
so that
$$S_V(f) \approx 4 k_b T R \tag{2}$$
which we traditionally call the "Johnson noise" formula.
If we short circuit the resistor as in the diagram where its ends are connected by a simple wire, then the current noise spectral density is (just divide by $R^2$)
$$S_I(f) = 4 k_b T / R \, .\tag{3}$$
Another way to think about this is that the resistor generates random current which is Gaussian distributed with standard deviation $\sigma_I = \sqrt{4 k_b T B / R}$ where $B$ is the bandwidth of whatever circuit is connected to the resistor.
Johnson noise keeps the system in equilibrium
Anyway, the point is that the little resistor in the machine actually generates random currents in the wire!
These little currents cause the rod to twist back and forth for exactly the same reason that the twists in the rod induced by air molecules crashing into the paddles caused currents in the resistor (i.e. Faraday's law).
Therefore, the thermal noise of the resistor shakes the paddles and heats up the air.
So, while heat travels from the air on the left side to the resistor on the right, precisely the opposite process also occurs: heat travels from the resistor on the right to the air on the left.
The heat flow is always occurring in both directions.
By definition, in equilibrium the left-to-right flow has the same magnitude as the right-to-left flow and both sides just sit at equal temperature; no entropy flows from one side to the other.
Fluctuation-dissipation
Note that the resistor is both dissipative and noisy.
The resistance $R$ means that the resistor turns current/voltage into heat; the power dissipated by a resistor is
$$P = I^2 R = V^2 / R \, . \tag{4}$$
The noise is characterized by a spectral density given in Eq. (1).
Note the conspicuous appearance of the dissipation parameter $R$ in the spectral density.
This is no accident.
There is a profound link between dissipation and noise in all physical systems.
Using thermodynamics (or actually even quantum mechanics!) one can prove that any physical system which acts as a dissipator of energy must also be noisy.
The link between noisy fluctuations and dissipation is described by the fluctuation-dissipation theorem, which is one of the most interesting laws in all of physics.
The machine originally looked like it moved entropy from the left to the right because we assumed the resistor was dissipative without being noisy, but as explained via the fluctuation-dissipation theorem this is entirely impossible; all dissipative systems exhibit noisy fluctuations.
P.S. I really, really like this question.
Best Answer
The second law holds on average for systems of any size, large or small. If you have an isolated contraption containing just a few atoms, and you run it through some procedure (maybe as simple as waiting 5 seconds, or maybe more complicated), there is some probability that the atoms will wind up in a lower-entropy configuration at the end of the procedure than the start. That's what your professor was referring to.
However, the probability of random entropy reduction is not high, and certainly not 100%. The key point is that the average change in entropy, upon many repetitions of the procedure, cannot be negative for any procedure.
When people hear this, they get an idea: "I'll run the procedure 100 times, and check each time whether or not the entropy got randomly lowered, and somehow I'll only use that 1 successful run to power the perpetual motion machine while throwing away the 99 unsuccessful runs." Unfortunately, it doesn't work that way! One would refer to this whole process (involving 100 runs of the original procedure) as being just one run of a different (more complicated) procedure, which also now involves extra effort / energy to check whether or not the entropy was lowered each time. (Otherwise you wouldn't know which of the 100 runs to use). That checking process creates enough entropy to undo the benefit of repeated runs.
This kind of stuff is commonly discussed under the heading of Maxwell's Demon.
If you have a small magnet diffusing around by brownian motion, it will indeed transfer energy to a stationary metal nearby via eddy currents. Unfortunately, the reverse process happens at exactly the same rate: The electrons in that metal randomly jiggle, creating currents that create magnetic fields that push on the diffusing magnet, thus transferring energy from the metal to the magnet. The total energy transfer rate is equal in both directions, or more specifically, as long as both parts start at the same temperature, they stay at the same temperature.
UPDATE WITH MORE DETAILS
A better way to define entropy is to say "We don't know exactly what the microstate (microscopic configuration) of a system is, instead our best information is that there a probability distribution of possible microstates. Then the entropy is $S = k_B \sum_n p_n \log p_n$ where $p_n$ is the probability of microstate $n$.
Note that entropy is observer-dependent, in the sense that one observer may have more information about the probability distribution than another. In more concrete terms, a system might be disordered to one observer, but a different observer knows a "magic recipe" for undoing that disorder. For example, I could create a seemingly-unpolarized beam of light by switching the polarization of a laser every nanosecond according to a pseudo-random sequence. I know the sequence, and therefore I can use a waveplate to get back a perfectly polarized beam with no intensity losses. But for somebody who doesn't know my pseudo-random sequence, the beam really needs to be treated as unpolarized, and they cannot polarize it without losses, according to the 2nd law.
A configuration of a small number of molecules might randomly become "more ordered" in some sense, but that doesn't mean it has a lower entropy. Until you measure it, you don't know that it became more ordered, and therefore you cannot make use of that order. All you have is a probability distribution for what the microstate is. As time passes the probability distribution changes and the entropy that you calculate from it either stays the same or goes up. Well, in a certain sense, it does not go up by Liouville's theorem in classical mechanics or unitarity in quantum mechanics ... but it often turns out that you wind up with useless information about the microstate, i.e. information that cannot be translated into a "magic recipe" for undoing the apparent disorder as in the polarization example above. In that case, you might as well just forget that information and accept a higher entropy. See this question.
When you make a measurement, there is a probability distribution for the possible measurement results and the possible microstates following the measurement. Some of the measurement results may leave you with a low-entropy configuration (you pretty much know what the microstate is from the measurement). But if you appropriately average over all possible measurement results, the overall entropy increases on avearge as a result of the full measurement process.