Using the 'momentum theorem' $F\Delta t=\Delta p$—（1）， and the weighing scale (at time $t$) results from two contributions:
One is the weight of the part of the chain that lie on the machine, which is equal to $\rho \frac{gt^2}{2}g$—（2）, where $\rho $ is the line density of the chain, $\frac{gt^2}{2}$ is the length of the lying part of the chain with $g$ the acceleration of gravity.
The other contribution comes from the force $\Delta F$ that acts on the falling chain to stop it. Specifically, imagine that at time $t$, the contacting point A on the chain will change to B after a very 'small' time interval $\Delta t$, and during this progress, the 'small' part of the chain (with length $v_t\Delta t$, where $v_t=gt$ is the velocity of the point A at time $t$) that connects A and B was stopped by the force $\Delta F$ which comes from the machine. Thus, apply Eq.(1) to this 'small' part, we have $(\Delta F-\Delta m\cdot g)\Delta t=\Delta m\cdot v_t$, where $\Delta m=\rho \cdot v_t\Delta t$ is the mass of the 'small' part, then we arrive at $\Delta F=\rho v_t^2=\rho g^2t^2(\Delta t\rightarrow0)$—（3）.
Finally, the weighing scale (at time $t$) $=Eq.(2)+Eq.(3)=\frac{3}{2}\rho g^2t^2$, which is 3 times of Eq.(2)—the weight of the lying part of the chain on the machine.
Remark: Referring to what confusing you in your last paragraph, one key difference between "you stand on the scale" and "the different part of the chain standing on the scale" is that the former is static while the latter is moving downward with acceleration of gravity $g$. Thus, back to your question, at any instant time $t$ the reading of the scale will not depend on the length(hence the mass) of the falling part of the chain.
In the elevator scenario, the elevator frame is getting accelerated; hence, the when you draw the free-body diagram, with respect to the elevator, the pseudo force acts downwards (opposite to the direction in which the frame is getting accelerated). Hence, the apparent weight increases as the pseudo force gets added up with the weight of the person.
Suppose the acceleration of the elevator is $a$ and the mass of the body is $m$, then the apparent weight of the body in the elevator frame is -
$$
N = m(a + g)
$$
In the second scenario, the buoyant force acts in the upward direction, because the buoyant force is always directed against the pressure gradient i.e, the direction in which pressure decreases. (Much like an electric field directed in the direction where the potential decreases)
Of course, the buoyant force exerted is equal to the weight of the fluid displaced by the body (Which is the Archimedes principle); but -
Drawing the FBD in the second case yields the weight of the body acting downwards, and the buoyant force acting upwards. This results in the weight decreasing (since the buoyant force is subtracted from the weight, not added up with it), and not increasing.
Say, the buoyant force acting on the body is $B$ and the actual weight is $W$, the net weight of the body (acting in the downward direction) then would be -
$$
W' = W - B
$$
Which is why the apparent weight of the body in the liquid decreases.
(This is considering that the density of the body is greater than the density of the liquid, in the case where it is opposite (the body doesn't sink; but floats partially), the signs of $W$ and $B$ are swapped and the net force is acting in the upward direction. In another scenario where the the weight of the the body is equal to the buoyant force, the net force on the body then is zero, hence it floats being completely submerged)
Keep in mind that a body loses weight in a liquid which is equal to the weight of the liquid displaced by it/equal to the buoyant force.
As for the bonus question, look into the answer to this question -
https://physics.stackexchange.com/a/296537/134658
Best Answer
Generally, a scale will measure the normal force it supplies to the object resting on it. In the special case where the scale is stationary (as it appears in your picture), this is equal to $mg$, or the weight of the object.
If the system is accelerating, the normal force (and thus the reading of the scale) will increase or decrease appropriately. However, this normal force is no longer equal to the weight.