Intuition behind the scaling property of Fourier Transforms
Let $g(x) = f(ax)$.  Then, using a change of variables $y = ax$,
$$\begin{align*}
G(u) &amp;= \int_{-\infty}^{\infty} g(x)\exp(-j u x)\,\mathrm dx\\
&amp;= \int_{-\infty}^{\infty} f(ax)\exp(-j u x)\,\mathrm dx\\
&amp;= \frac{1}{|a|}\int_{-\infty}^{\infty} f(y)\exp(- j (u/a) y)\,\mathrm dy\\
&amp;= \frac{1}{|a|}F\left(\frac{u}{a}\right)
\end{align*}$$
not $\displaystyle \frac{1}{|a|}F\left(\frac{u}{|a|}\right)$ as you say (cf. Henry Gomersall's
comment on your question).
Now, if $f(x)$ is a periodic function with period $2\pi$, then its 
Fourier transform exists only in a generalized sense (involves impulses).
The Fourier series
representation $\displaystyle \sum_n c_n \exp(j nx)$ of $f(x)$
has Fourier coefficients
$$\displaystyle c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)\exp(-j n x)\,\mathrm dx,
 ~ -\infty &lt; n &lt; \infty$$ 
and these also show up in its Fourier transform 
which is given by
$\displaystyle F(u) = \sum_n c_n \delta(u -n)$. Note that
the inverse Fourier transform is
$$\int_{-\infty}^\infty F(u)\exp(jux)\,\mathrm du
= \sum_n \int_{-\infty}^\infty c_n \delta(u-n)\exp(jux)\,\mathrm du
= \sum_n c_n \exp(j nx) = f(x).$$
The function $g(x) = f(ax)$ of period $2\pi/a$ thus has Fourier transform
$$G(u) = \frac{1}{|a|}\sum_{n=-\infty}^{\infty} c_n \delta(u/a -n) 
= \frac{1}{|a|}\sum_{n=-\infty}^{\infty} c_n \delta\left(\frac{u-an}{a}\right).$$
But, $\delta(\alpha x) = \frac{1}{|\alpha|}\delta(x)$, and so with 
$\alpha = a^{-1}$, we have
$$G(u) = \sum_n c_n \delta(u - an)$$
whose inverse Fourier transform is
$$\int_{-\infty}^\infty G(u)\exp(jux)\,\mathrm du
= \sum_n \int_{-\infty}^\infty c_n \delta(u-an)\exp(jux)\,\mathrm du
= \sum_n c_n \exp(j nax) = g(x) = f(ax).$$

So, you are not missing anything. Scaling an ordinary
function scales both
the magnitude and the frequency axis of the Fourier transform, 
but when impulses are involved, scaling the argument of the
impulse leads to a magnitude scaling that exactly compensates
for the magnitude scaling imposed on the Fourier transform,
and leaves the Fourier coefficients unchanged.
frequency scaling also scales the amplitude