Solution:
(1) $h(x) = \sqrt{3}\cos(x + \frac{\pi}{6}) + 3\cos(\frac{\pi}{3} - x) = \sqrt{3}\cos(x + \frac{\pi}{6}) + 3\cos[\frac{\pi}{2} - (x + \frac{\pi}{6})] = \sqrt{3}\cos(x + \frac{\pi}{6}) + 3\sin(x + \frac{\pi}{6})$,
Therefore, the function $h(x)$ has a companion vector $\overrightarrow {OM} = (3, \sqrt{3})$,
Therefore, the collinear unit vector with $\overrightarrow {OM}$ is $\left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$.
(2) The companion function $f(x)$ of $\overrightarrow {OM}$ is $f(x) = a\sin x + b\cos x = \sqrt{a^2 + b^2}\sin(x + \varphi)$,
where $\cos\varphi = \frac{a}{\sqrt{a^2 + b^2}}$, $\sin\varphi = \frac{b}{\sqrt{a^2 + b^2}}$
When $x + \varphi = 2k\pi + \frac{\pi}{2}$, $k \in \mathbb{Z}$, i.e., $x_0 = 2k\pi + \frac{\pi}{2} - \varphi$, $k \in \mathbb{Z}$, $f(x)$ attains its maximum value,
Therefore, $\tan x_0 = \tan(2k\pi + \frac{\pi}{2} - \varphi) = \cot\varphi = \frac{a}{b}$,
Therefore, $\tan 2x_0 = \frac{2\tan x_0}{1 - \tan^2 x_0} = \frac{2 \times \frac{a}{b}}{1 - (\frac{a}{b})^2} = \frac{2}{\frac{b}{a} - \frac{a}{b}}$.
Let $m = \frac{b}{a}$,
then $\tan 2x_0 = \frac{2}{m - \frac{1}{m}}$, $m \in [0, \sqrt{3}]$,
Therefore, $\frac{1}{m} \geq \frac{\sqrt{3}}{3}$,
Therefore, $- \frac{1}{m} \leq - \frac{\sqrt{3}}{3}$,
Therefore, $m - \frac{1}{m} \in (0, \frac{2\sqrt{3}}{3}]$,
Therefore, $\tan 2x_0 \in (-\infty, 0) \cup \left[\sqrt{3}, +\infty\right)$.

Thus, the final answers are:
(1) The collinear unit vector with $\overrightarrow {OM}$ is $\boxed{\left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)}$.
(2) The range of values for $\tan 2x_0$ is $\boxed{(-\infty, 0) \cup \left[\sqrt{3}, +\infty\right)}$.