What S(m) = T(2^m) means is that S, T are two different functions which produce the same result while taking inputs as m and 2^m respectively.

Function S can be considered as an operator with two internal steps (otherwise, composition of functions):

1. S' operator: Input:m, Output:2^m

2. T operator(original function): Input:output of first part, Output: As defined originally.

Therefore then transitions are:

m->2^m->T(2^m) = S(m)

(m/2)->(2^(m/2))->T(2^(m/2)) = S(m/2)

What $S(m) = T(2^m)$ means is that $S$ and $T$ are two different functions which produce the same result while taking inputs as $m$ and $2^m$ respectively.

Function $S$ can be considered as an operator with two internal steps (otherwise, composition of functions):

1. $S'$ operator: Input:$m$, Output:$2^m$

2. $T$ operator(original function): Input:output of first part, Output: As defined originally.

Therefore then transitions are:

$$m\to 2^m\to T(2^m) = S(m)$$ $$\tfrac{m}2\to2^{m/2}\to T(2^{m/2}) = S(\tfrac{m}2)\,.$$