About negative energies: they set no problem:
On this context, only energy differences have significance. Negative energy appears because when you've made the integration, you've set one point where you set your energy to 0. In this case, you have chosen that $PE_1 = 0$ for $r = \infty$. If you've set $PE_1 = 1000$ at $r = \infty$, the energy was positive for some r.
However, the minus sign is important, as it is telling you that the test particle is losing potential energy when moving to $r = 0$, this is true because it is accelerating, causing an increase in $KE$:
let's calculate the $\Delta PE_1$ for a particle moving in direction of $r = 0$: $r_i = 10$ and $r_f = 1$:
$\Delta PE_1 = PE_f - PE_i = Gm(-1 - (-0.1)) = -Gm\times0.9 < 0$
as expected: we lose $PE$ and win $KE$.
Second bullet: yes, you are right. However, it is only true IF they are point particles: has they normally have a definite radius, they collide when $r = r_1 + r_2$, causing an elastic or inelastic collision.
Third bullet: you are right with $PE_2 = mgh$, however, again, you are choosing a given referential: you are assuming $PE_2 = 0$ for $y = 0$, which, on the previous notation, means that you were setting $PE_1 = 0$ for $ r = r_{earth}$.
The most important difference now is that you are saying that an increase in h is moving farther in r (if you are higher, you are farther from the Earth center).
By making the analogy to the previous problem, imagine you want to obtain the $\Delta PE_2$. In this case, you begin at $h_i = 10$ and you want move to $h_f = 1$ (moving in direction to Earth center, like $\Delta PE_1$:
$\Delta PE_2 = PE_{f} - PE_{i} = 1mg - 10mg = -9mg < 0$.
As expected, because we are falling, we are losing $PE$ and winning $KE$, the same result has $PE_1$
Fourth bullet: they both represent the same thing. The difference is that $gh$ is the first term in the Taylor series of the expansion of $PE_1$ near $r = r_{Earth}$. As exercise, try to expand $PE_1(r)$ in a taylor series, and show that the linear term is:
$PE_1 = a + \frac{Gm(r-r_{earth})}{r_{earth}^2}$.
Them numerically calculate $Gm/r_{earth}^2$ (remember that $m=m_{earth}$). If you haven't made this already, I guess you will be surprised.
So, from what I understood, your logic is totally correct, apart from two key points:
energy is defined apart of a constant value.
in the $PE_1$, increase r means decrease $1/r$, which means increase $PE_2 = -Gm/r$. In $PE_2$, increase h means increase $PE_2=mgh$.
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 weight of an object is the force given by Newton's second law:
$$ F = ma $$
As you say the mass is constant, and if an object is neither accelerating up nor down the acceleration is just the gravitational acceleration $g \approx 9.81 \text{ms}^{-2}$. Then we get the familiar equation for the weight:
$$ W = mg $$
However suppose you are doing a loop. If you've ever done this you'll remember that your weight increases at the bottom of the loop and decreases at the top of the loop. That's because at the bottom of the loop the plane is accelerating upwards and at the top of the loop the plane is accelerating downwards. If we call the plane's acceleration $a_p$ then the total acceleration is:
$$ a = g + a_p $$
and the weight becomes:
$$ W = m(g + a_p) $$
At the top of the loop the plane is accelerating downwards so $a_p \lt 0$ and that means $W \lt mg$ i.e. your weight is reduced.
Though I've never done it I'm told it is possible to pull a loop so tight that you experience negative acceleration at the top of the loop i.e. the plane's acceleration downwards is greater than the gravitational acceleration upwards and your weight becomes negative. I would guess this is what the post on the Aviation SE means.