The input to the multiple stage machine was not 2. It is the empty string, $\varepsilon$.
This is fed into the machine $T_n$ which has $n$ states and outputs $n$ in unary.
At the next stage, $n$ (in unary) is fed to the doubling machine, which outputs $2n$ in unary.
This is then fed to the presumed BB machine, which outputs $BB(2n)$ in unary.
How many states does the compound machine have?
$T_n$ has $n$ states
double has $d$ states for some constant $d$
BB has $b$ states for some constant $b$
and let us say $C$ states for gluing everything together (again for some constant $C$).
All by all $n+d+b+C$ states. And on the empty input string it outputs $BB(2n)$ in unary,
forcing it to work at least that many steps.
Take $n$ for which $d+b+C<n$, implying $n+d+b+C<2n$. Then with fewer than $2n$ states, the compound machine
worked more than $BB(2n)$ steps on the empty input string, and stopped.
This contradicts the definition of $BB(\cdot)$.