Why can't I separate the fraction inside a fraction to the outside I know when I have something of this form
\begin{equation}
{a} \frac{1}{\alpha} \frac{s + \frac{1}{T}}{s + \frac{1}{\alpha T}}
\end{equation}
I multiply out the fractional part,  α with the bottom to get
\begin{equation}
{a} \frac{s + \frac{1}{T}}{\alpha s + \frac{1}{T}}
\end{equation}

But, if I instead have
\begin{equation}
{a}  \frac{\frac{1}{\alpha} ( s + \frac{1}{T} )}{s + \frac{1}{\alpha T}}
\end{equation}
with the fractional part 
\begin{equation}\frac{1}{\alpha} \end{equation} in the numerator instead.

I know I can't take the fractional part out of the equation in this second instance. But, I don't know why that is. I know this is a fairly basic rule, but I haven't done algebra in quite a while. So, I wanted to ask for clarification.
 A: 
I know I can't take the fractional part out of the equation in this second instance

What?  Yes you can.  (Technically it's "expression" and not "equation.")
In general, $\dfrac{AB}C = A\dfrac BC$, as long as $C \ne 0$ since otherwise these expressions are undefined.
In your case:
$$
a \ \frac{\frac1\alpha ( s + \frac1T )}{s + \frac1{\alpha T}}
 = a \ \frac1\alpha \frac{ s + \frac1T}{s + \frac1{\alpha T}}
$$
A: if you have this 
\begin{equation}
{a}  \frac{\frac{1}{\alpha} ( s + \frac{1}{T} )}{s + \frac{1}{\alpha T}}
\end{equation}
then it is exactly the same as
\begin{equation}
{a} \frac{1}{\alpha} \frac{s + \frac{1}{T}}{s + \frac{1}{\alpha T}}
\end{equation}
And it is a matter of asociation: Lets call $b:= \frac{1}{\alpha},\ \ c:=s + \frac{1}{T},\ \ d:=s + \frac{1}{\alpha T}$. Then, rewriting the fractions as divisions, we have that 
\begin{equation}
 \frac{\frac{1}{\alpha} ( s + \frac{1}{T} )}{s + \frac{1}{\alpha T}} = (b.c):d =b.(c:d) = \frac{1}{\alpha} \frac{s + \frac{1}{T}}{s + \frac{1}{\alpha T}}
\end{equation}
