Question about proving that Rado's function is non-computable
Thanks for the clarification! I really misinterpreted your question.

Okay, so we have the computable function $f$, the also computable function $F$ that is based on $f$, and the Turing machine $M_F$ with the following behavior:


  when started with a tape with $y$ $1$'s, writes a block of $F(y)$ $1$'s to the right, separated by at least one blank


Note that I used $y$ instead of $x$. So when our TM starts with a tape like

$$\Delta\underbrace{1\ldots1}_{\times y}\Delta$$

the resulting tape will be 

$$\Delta\underbrace{1\ldots1}_{\times y}\Delta\underbrace{1\ldots 1}_{\times F(y)}\Delta$$

and we say that $M_F$ uses $n$ states. Note that $n$ is independent of $y$. The number of states doesn't depend on the input of the TM.

Now the machine $M$ does the following:


Start with blank tape: $\Delta$
Write $x$ $1$'s on tape: $$\Delta\underbrace{1\ldots1}_{\times x}\Delta$$
Now mimic $M_F$ with these $x$ $1$'s as a start, i.e. $y=x$: $$\Delta\underbrace{1\ldots1}_{\times x}\Delta\underbrace{1\ldots 1}_{\times F(x)}\Delta$$
Now we mimic $M_F$ again, but this time with the $F(x)$ $1$'s as a start, i.e. $y=F(x)$: $$\Delta\underbrace{1\ldots1}_{\times x}\Delta\underbrace{1\ldots 1}_{\times F(x)}\Delta\underbrace{1\ldots 1}_{\times F(F(x))}\Delta$$


So when we mimic $M_F$ a second time, we simply ignore the $x$ $1$'s at the leftmost of the tape and pretend they are not there. I think the text is implying that executing $M_F$ doesn't moves it too much left and that is understandable, since this behavior can be simulated by using a one-side bounded TM which is equivalent to our usual TM. So by mimicking $M_F$ two times, we would need about $2n$ states for that.