What part of the cylinder displaces some volume from the second liquid (which is used for the buoyant force)? Consider a container filled with two liquids, of densities $\rho_1$ and $\rho_2$, such that $\rho_2 < \rho_1$. Now imagine placing a cylinder with the top base removed, of density $\rho,\space \rho_2<\rho<\rho_1$, such that the cylinder floats between the liquids. The second liquid is barely filling the cylinder, whose interior diameter is $l_1$ and whose exterior diameter is $l_2$. An illustration can be found below:

Now I am a little bit confused about the buoyancy force in liquid $2$: I thought I could express it as (assuming the height of the cylinder in liquid $2$ is $h$):
$$F_{B2} = (S_2-S_1)h\rho_2g=(\pi\frac{l_2^2}{4}-\pi\frac{l_1^2}{4})h\rho_2g=\\=\frac{\pi h\rho_2g}{4}(l_2^2-l_1^2)$$
Because the liquid actually enters the cylinder and only the walls displace a volume of liquid. However, the authors of the problem and my teacher claim that:
$$F_{B2}=\frac{\pi l_2^2h\rho_2g}{4}$$
Which means that the entire part of the cylinder that lies in liquid $2$ actually displaces a certain part of its volume, rather than the walls only. I have strong reasons to believe that I am wrong, because this is a tiny sub-part of a problem from a national contest, and the official solutions claim that the latter is correct. Which version is correct and why? What part of the cylinder displaces some of the second liquid, the walls or the cylinder itself as a whole (well, the part of them that lie in the second liquid) and why?
 A: Where your confusion lies, I think, is whether the volume of fluid 2 in the cylinder should be counted as "displacing" some volume of fluid 2 in the larger container, or whether it shouldn't be counted since that fluid would already "be there" if the cylinder wasn't there.  The answer, I think, is that the authors' answer is the conventional way to do things;  but if you were careful, you could do it your way too.
Let's consider a simpler example:  a barrel full of water is submerged in a tank of water, hanging from a spring scale above the tank.  I'll ask two questions:  what is the buoyant force on this tank?  And what is the reading of the scale?
Under the conventional definition, one would say that the buoyant force on the barrel is
$$
F_B = (V_\text{barrel} + V_\text{water}) \rho g,
$$
where $V_\text{barrel}$ and $V_\text{water}$ are the volume of the container and the volume of the water inside the tank, respectively, and $\rho$ is the density of water.  This buoyant force would be directed upwards.  Similarly, the weight of the tank would be
$$
F_G = (m_\text{barrel} + m_\text{water}) g = m_\text{barrel}g + V_\text{water} \rho g, 
$$
directed downwards.  The net upward force applied by the scale to keep the barrel from moving (which would be its reading) would then be
$$
F_S = F_G - F_B = m_\text{barrel}g - V_\text{barrel} \rho g.
$$
Now let's do it your way, and treat the water inside the barrel as part of the water in the tank.  In this case, the only volume of water that is displaced is the volume of the barrel itself, and so the buoyant force is
$$
F_B = V_\text{barrel} \rho g
$$
upwards.  The weight, meanwhile, is also just the weight of the barrel;  remember that we're viewing the water inside the barrel as part of the tank, so we don't have to count it:
$$
F_G = m_\text{barrel} g.
$$
The net force on the scale is therefore
$$
F_S = (m_\text{barrel} - V_\text{barrel} \rho) g,
$$
exactly as it was before.
All of this stems from the fact that the buoyant force on an object is, by definition, the weight of the displaced fluid.  If we fill some or all of that volume by the displaced fluid, we can either view this as increasing the weight of the object (while keeping the displacement constant) or decreasing the displacement of the object and thereby the buoyant force (while keeping the weight constant.)  Either way of looking at it leads to the same net change in the force on the object.
