I agree that the attempt at explanation offered by Veritasium (Derek Muller) is unconvincing.
Now, it is not straightforward to make the motion of lifting the non-spinning wheel the same as lifting the spinning wheel. In the case of the spinning wheel the rate of precession is a given, that dictates the rythm of the lift. Derek keeps his feet on the ground, he is not moving his body to match the orientation of the bar, so the weight shifts over the muscle group of his shoulder as he goes through the lift.
Indeed the force required to lift the spinning wheel must be the same as the force required to lift the non-spinning wheel.
(Experimental verification of that is actually rather tricky. I've seen a video of someone who created a tabletop setup, with normal gyroscope. His measurement reading swung up and down a bit, he had to average. Derek tries it with that giant wheel, but the setup he uses isn't stable enough.)
Anyway, my best guess is that in lifting the spinning wheel Derek's arm is moving relative to his body, shifting the load from muscle to muscle, whereas in the other lift he is just killing one particular muscle. He really should have tried to replicate that flow of motion with a barbell for one-handed lift.
I just tried it with a barbell for one-handed lift. Lifting with my right arm I maintained a clockwise rotation of the barbell. Lifting the barbell was doable that way. Then I tried lifting with my arm at a fixed angle to my body. That was significantly harder. My best guess: at different horizontal angles to your body your upper arm has a different vertical angle of optimal leverage. Presumably that depends on the placements of muscle attachments. Apparently it so happens that the rotation takes a path where all along the lift you have good leverage. Conversely, if you try to lift with your arm at a constant horizontal angle to your body you will inevitably hit a vertical angle with poor leverage.
Additional remarks (3 hours after initial submission of this answer):
Derek Muller mentions: if you have a gyroscope wheel with the spin axis in the horizontal plane (hence you get a torque from gravity), then when you push to give the wheel a surplus above the natural precessing motion the wheel will climb. (Conversely, when you push against the natural precessing motion the wheel will descend.) Derek offers that as a possible explanation for why it feels to hem that lifting the spinning wheel is less hard.
This suggested explanation doesn't work. The reason for that: at the very instant you stop pushing this happens: the precession rate of the wheel goes back to the natural precession rate.
Also, if Derek would not push at all, if he would simply release gingerly the wheel will go into the natural precession rate on its own.
In all: Derek's pushing arm can assist the lifting arm a little, but that assistence ceases when the pushing arm releases. At that point the wheel is still below the level of Derek's elbows.
Incidentally, this error doesn't stand on its own
There is another video by Derek about gyroscopic precession with an error.
Video about gyroscopic precession 2:47 into the video
what happens if I only let go after I've already spun up the bicycle
wheel well in that case the bicycle wheel would already have angular
momentum this way and so a torque pushing that way actually swings
this angular momentum round that way
(To avoid misunderstanding: Derek isn't the only one to make this claim; the claim does not originate with him.)
The suggestion is that the torque from gravity is being redirected, causing a precession instead of downward acceleration. The problem is: if that would be the case then the precession would speed up; a sustained force causes acceleration. But as we know: given a particular spin rate of the wheel, and a particular torque, there is a corresponding constant rate of precession. So: the suggestion that the torque from gravity is being redirected violates the laws of motion.
For a discussion of the mechanics of gyroscopic precession see my 2012 answer, the question is titled: What determines the direction of precession of a gyroscope?
Best Answer
The quick answer is that gyroscopes don't do what you suggest, and indeed don't do anything not expected from the well established laws of mechanics. Because gyrosocopes are so widely used in inertial guidance systems their behaviour has been measured to exquisite precision e.g. in the Gravity Probe B satellite. Any effect of the type you describe would be easily measured.
Having said this, every student is baffled by gyroscopes when they first encounter them, and that certainly includes me. However the equations that describe their motion are very simple. I think the problem is that the gyroscope equivalent of Newton's first law is:
$$ \vec{\tau} = \frac{d\vec{I}}{dt} $$
where $\vec{\tau}$ is the torque and $\vec{I}$ is the angular momentum. The big difference from linear motion is that the torque is given by:
$$ \vec{\tau} = \vec{r} \times \vec{F} $$
where $\vec{F}$ is the force and $\vec{r}$ is the vector from the centre of rotation to the point where the force is applied, and the $\times$ is a cross product. The cross product produces a vector that is at right angles to the two vectors in the product, which means the torque is a vector at right angles to the applied force, and therefore the change in angular momentum is at right angles to the applied force.
It's the fact that gyroscopes respond in a different direction to the force you apply that makes their behaviour seem so counterintuitive. But I must emphasise that it's our intuition that is the problem, not the gyroscope. Once you've mastered the maths, the behaviour of gyroscopes is very simple and entirely predictable.
Response to comment:
The rocket on the left has the gyroscope rotating about it's centre i.e. the gimbal in which it's mounted rotates about the centre. The dotted line shows the thrust provided by the rocket motor. In this case the net torque is zero because the same force is applied to both ends of the gyroscope and the torques cancel out. This rocket would rise normally.
In the right hand drawing the gyroscope is pivoting about it's end. In this case there is a net torque and the rocket would feel a force rotating it out of the plane of the diagram. If the angular momentum of the gyroscope were large enough (it wouldn't be in real rockets) the rocket would indeed rotate round and crash into the ground.