It has to do with the principal atomic number of those f orbitals. Some quick review (where ~ means "is related to in a monotonic, increasing fashion")

n = Principle Atomic Number ~ energy,size of orbital
l = Total Angular Momentum Number ~ abs(ang m.)^2

n is 1,2,3,...
l is 0,1,2,... (0=S 1=P 2=D 3=F ...)

Now you can tell that these two should be related by thinking about the classical "solar system" electron/nucleus model: if an electron has lots of angular momentum that means it must have lots of energy. Careful: if it has no angular momentum it could still have lots of energy if it's going straight at or away from the the nucleus (angular momentum cares about velocity perpendicular to the line between two objects; energy cares about the absolute velocity). This idea turns into an inequality we get by solving the Schrodinger equation for a H atom to find eigenstates which we parameterize on the quantum numbers:

l < n

also we can figure out from the H-atom Schrodinger equation

Increasing n only (going down a column) increases energy

The different "blocks" of the periodic table are named after the l quantum number because it is instrumental in determining the shape (as opposed to size) of the orbital. S is the 2 cols on the left, D is 10 cols in the middle, P is 6 on the right, and F is down below (it should actually be between S and D blocks but that wastes space on posters).

What does the inequality (and the intuition behind the inequality) tell us about these blocks? Let's go one by one

S block. l=0. 0<n. n>=1. S orbitals have "size 1" or larger
P block. l=1. 1<n. n>=2. P orbitals have "size 2" or larger
D block. l=2. 2<n. n>=3. D orbitals have "size 3" or larger
F block. l=3. 3<n. n>=4. F orbitals have "size 4" or larger

This sort of looks like larger l means larger orbitals, but not so fast! They fill in order of energy level, not momentum (low energy = tightly bound to nucleus, deep in its electrostatic well). If we fill orbitals one by one, we know the first P orbital will have size 1 but that doesn't tell us anything about the S orbitals that come before it, and indeed there is an S orbital of size 2 that comes before the first P orbital of size 1. Just read off the periodic table in order of increasing (proton count = electron count) to see what I mean.

OK, now we jump to the F block. We know the first f orbital we see has size 4 -- it is the outermost orbital of Lanthanum. But what is the preceding orbital, the one with slightly less energy (slightly more tightly bound to the nucleus)? It's the outermost orbital of Barium, and it's an S orbital of size 6.

S orbital of size 6 is bigger than a P orbital of size 4 :-)

This is counterintuitive because it breaks the "bigger orbitals have more energy / are less tightly bound" rule. The solution to the paradox is that the part of the orbital (probability density function) that contributes most to the energy is the part near the nucleus where the electron is far down the rabbit hole, so to speak. While the outermost part of 4f is smaller than that of 6s, the inner part of the 4f has pulled away from the nucleus while the 6s still sits squarely on top of it. 6s gets out to play with other atoms more than 4f but in terms of energy it more than balances that out by spending time really close to the nucleus.

Hope this helps!

EDIT: fixed an off by one bug on the inequalities