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
Archimedes' principle isn't directly applicable here, since there's nowhere for the water to be displaced to. If you filled a thinner, weightless beaker with the oil and tossed that into the original beaker, you will indeed see water displaced. But in your example, the water can't even be displaced (where would it go?) so you don't see displacement. The entire system of oil and water can be treated as one.
Here's another way to think about it. In the figure on the left, imagine cutting the water in half such that the top layer is resting on the bottom one. The top layer is clearly in equilibrium, or it would move. But gravity is acting on it, so there must be a force cancelling gravity. The bottom layer isn't displaced either, so buoyancy isn't the source of this force. Instead it's coming from the internal pressure of water, which is preventing the bottom layer from being compressed.
Best Answer
The argument sounds perfectly reasonable.
Consider arbitrary parcel of fluid in equilibrium. It exerts downward force equal to it's weight on the surrounding fluid, and it does not move. Therefore according to the second law of motion, the downward force must be balanced by upward force of equal magnitude, the buoyant force (otherwise it would start to move, contradicting the equilibrium assumption).
The buoyant force is exerted by the fluid surrounding the parcel. Therefore if we replace the parcel with something else, there is no reason for that force to change.