Fourier Transform Time Scaling Property using f (frequency) or w (omega - angular frequency)
If you're getting hung up on these extra \$2\pi\$ scaling factors, then you're probably missing the entire concept of the time scaling properly.  Basically, if you shrink in time you expand in frequency, and if you expand in time you shrink in frequency...no matter how the frequency was initially represented.  Think of how a \$rect()\$ function in one domain becomes a \$sinc()\$ in the other.  The extra \$1/|a|\$ scale factor is to compensate for conservation of energy between each domain (see Parseval's Theorem).
When discussing transform properties, the original transform is already defined and you're only focusing on how the property modifies that original transform.  So it's probably easier to follow if you first write out the default transform representation, followed by one with the property applied.  If these properties are conveyed in a table, then they usually put the default/reference transform at the top so you can easily compare against it.
In this specific case, with time scaling, it modifies your existing representation of frequency (whatever it was).  Here are some examples using different frequency symbols/representations that hopefully clarify this for you.
$$
\begin{align*}
&amp;\text{Fourier Transform:} &amp;&amp;\mathcal{F}\{g(t)\} = G(f) \implies \mathcal{F}\{g(at)\} = \frac{1}{|a|}G\left(\frac{f}{a}\right) \\ \\
&amp;\text{Laplace Transform:} &amp;&amp;\mathcal{L}\{g(t)\} = G(s) \implies \mathcal{L}\{g(at)\} = \frac{1}{|a|}G\left(\frac{s}{a}\right) \\ \\
&amp;\text{Poo Transform:} &amp;&amp;\mathcal{P}\{g(t)\} = G(poomega) \implies \mathcal{P}\{g(at)\} = \frac{1}{|a|}G\left(\frac{poomega}{a}\right)
\end{align*}
$$