Why is the weight ignored in buoyant force derivations? why is it that the weight that the object makes is totally disregarded in these derivations of the buoyant force? Shouldn't that F up be the weight of the water in the top + the weight of the object? But instead it's calculated as if the object was a portion of water, why?

Is this related to the fact that there's water around the object? But what if there isn't, like in the case below (ignore friction), where there is a pink object that has the same width as the container. Should the pressure in that point still be calculated as pw g h? Or it should be (pw g h a) / a (weight of portion of water on top divided by area) + mg/a (weight of the object divided by area)?

 A: Buoyant force and Archimedes. Buoyant force is defined as the resultant of the stress acting on the surface of a body immersed in a fluid at rest.
The fluid surrounding the immersed body "doesn't know anything" about the material of the body. The equilibrium of the fluid holds in the very same way it would hold if the immersed body is replaced by the fluid itself. Thus the weight of the immersed volume must be equal to the weight of the same volume of the fluid.
But this is only the force acting on the body, because of its interaction with the surrounding fluid. When you write equilibrium equations of the body, you need to write that the sum of the forces acting on the body (buoyant force pushing upwards and its weight pushing downwards)
$0 = F^{bou} - F^{wei} = \rho_f V_{imm} g - m g$
Fluid column example. If you had three regions of fluids, you could recursively apply Stevino's law, starting on a surface where you know the pressure, and assuming no pressure jump across the interface between two fluids (neglecting the small contribution of surface tension).
Here with a solid, you need to rely on equilibrium condition for the solid (if we're considering a steady state), while for the fluid you can use Stevino's law. If the upper fluid has a free surface, and you know the pressure on it, this is the boundary condition we need for integration. To be clear,

*

*fluid 1: starting from the free surface at $P_O$, pressure is a function of depth as
$P(z) = P_0 + \rho_1 g z \quad , \quad z \in [0, z_1]$
so that the pressure acting on the upper surface of the solid is $P_1 = P_0 + \rho_1 g z_1 $


*solid: the equilibrium equation of the solid reads
$0 = -mg + (P_2-P_1)A$,
where $P_2 = P_1 + \dfrac{mg}{A}$ is the pressure acting on the lower surface of the solid.


*fluid 2: pressure in the lower portion of fluid is governed by Stevino's law
$P(z) = P_2 + \rho_2 g z \quad , \quad z \in [0, z_{bottom}]$,
measuring the depth starting from the lower surface of the solid.
