Question about proving that Rado's function is non-computable

Question

I am currently following the proof, found here, that Rado's function $$ \Sigma (n) = \max \{ \text{# of 1's that may be written to a tape by an n-state turing machine} \} $$ is non-computable. Within this proof, they say that there exists a Turing machine $M_F$, which we say has $n$-states, that writes $F(x)$ 1's to the tape.

However, they then go on to imply that there exists a Turing machine with $n$ states which writes $F(F(x))$ 1's to the tape! Surely, since $F(x) > x$ this should require more than $n$ states - since more states $\iff$ more 1's?

EDIT: For added clarity, I will now explain why I have interpreted this to mean that both $F(x)$ and $F(F(x))$ have $n$ states.

This proof first defines a Turning machine $M_F$ which has $n$ states and writes $F(x)$ 1's to the tape before halting. It then goes on to describe a (trivial) $x$-state Turing machine $M$ which "on input $\Lambda$" writes $x$ 1's to the machine's tape.

The proof then goes on to claim that $$ \Sigma (x + 2n) \geq x + F(x) + F(F(x)) $$ which means that there exists a $x + 2n$ state Turing machine which prints $x + F(x) + F(F(x))$ 1's to the machines tape.

I am assuming that these $x + F(x) + F(F(x))$ 1's are written as follows:

1. The $x$ state Turing machine $M$ prints $x$ 1's to the tape.
2. Next, $F(x)$ 1's are written to the tape by the $n$ state Turing machine $M_F$
3. So far, we have written $x + F(x)$ 1's and $x+n$ states have been used to do so. Comparing this to the fact that $x + F(x) + F(F(x))$ 1's may be written using $x + 2n$ states, they are implying that there must exist a Turing machine which writes $F(F(x))$ 1's using $n$ states

I am probably interpreting this incorrectly, but I don't know why?