Solution:
Setting $h(x)=f(x+1)-1$, it is easy to see that the conditions for $f(x)$ are equivalent to $h(x)=a x$ for any $x \in [1,2)$ and $h(h(x))=-2 x$ for any $x \in \mathbb{R}$. Then $h(-2 x)=h(h(h(x)))=-2 h(x)$; in particular, $h(0)=0$.

It follows by induction that $h\left(4^{n} x\right)=4^{n} h(x)$ and hence $h(x)>0$ for $x \in \left[4^{n}, 2 \cdot 4^{n}\right)$, where $n$ is an arbitrary integer. Since $0>-2 x=h(h(x))=h(a x)$ for $x \in [1,2)$, then $[a, 2 a) \subset \left[2 \cdot 4^{k}, 4^{k+1}\right)$ for some integer $k$. Therefore $a=2 \cdot 4^{k}$.

Conversely, if $a$ has this form, then it is easy to check that the function
$$h(x)=\left\{\begin{aligned} a x, & \quad x \in \left[4^{n}, 2 \cdot 4^{n}\right) \\ -\frac{2 x}{a}, & \quad x \in \left[2 \cdot 4^{n}, 4^{n+1}\right) \\ 0, & \quad x=0 \\ a x, & \quad x \in \left(-4^{n+1},-2 \cdot 4^{n}\right] \\ -\frac{2 x}{a}, & \quad x \in \left(-2 \cdot 4^{n},-4^{n}\right] \end{aligned}\right.$$
where $n$ runs over all integers, has the desired properties. One can easily show that this is the only function with the above properties.