You are right that no function can have compact support both before and after Fourier Transformation.
To realize this, imagine we do something that chops the function to it's support.
$$f_{chop}(t)= f(t)\cdot (H(t_0+t) \cdot H(t_1-t))$$
where
$$B_{t_0,t_1}(x) = (H(t_0+t) \cdot H(t_1-t))$$
is a box function from $t_0$ to $t_1$:
Now we can derive ( usually done in any first course in Fourier Analysis or Transform Theory ) the Fourier transform of this Box:
$$\mathcal{F}\{B_{t_0,t_1}\} = \text{sinc}((t_1-t_2)w) = \frac{\sin(w(t_1-t_2))}{w(t_1-t_2)}$$
This sinc function does not have compact support, since there is no $w$ it becomes zero for (except exactly each period of the sine).
Now the crucial part, the convolution theorems:
$$(f*g)(t) = \mathcal{F}^{-1}(\mathcal{F}\{f\} \cdot \mathcal{F}\{g\})$$
$$f(t)\cdot g(t) = \mathcal{F}^{-1}(\mathcal{F}\{f\} * \mathcal{F}\{g\})$$
Convolution in one domains gives multiplication in the other, and vice versa.
This means if something strips a function of it's infinite support in one domain (multiplying with a box), then it automatically becomes convolution with a infinite supported sinc function in the other domain.
Best Answer
I think of it as follows:
Now concerning your formula : $$ \mathcal{F}\left(t\mapsto f(at)\right)=\frac{1}{\mid a \mid}\hat{f}\left(\frac{\xi}{a} \right) $$