The force required to push an object into water increases as the object submerges, i.e. as the amount of water the object displaces steadily increases. But I think if you do the experiment carefully you will find that, once the object is fully submerged, the force required should be almost constant.
Thereafter, many objects get easier to push down with increasing depth, as the water pressure crushes them and they therefore displace less water. Wetsuits, for example, become greatly less buoyant with depth for this reason, which is why divers usually wear a buoyancy compensator.
At extreme depths, if something is less compressible than water, it will become harder to push down owing to the increasing density of water with depth. Factors such as this are important in the design of deep sea submersibles and bathyscaphes such as Alvin and the Trieste.
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
The fluid does not really exert an upward force on a body. It exerts a force everywhere on the body normal to its surface.
[Adapted from Hyperphysics]
In a gravitational field, the pressure increases with depth, so those normal forces which would otherwise cancel, end up summing to an upward force vector. If the tank were placed on, say, a centrifuge, where the local acceleration pointed outward and caused a sideways pressure gradient, the object would likewise "float" sideways.