How to calculate wavenumbers of normal modes from the eigenvalues of the Cartesian force constant matrix?
Short answer: (1) mass weight each force constant (see below). (2) Diagonalize the matrix. (3) Convert the eigenvalues ($\{ k_1^{\prime\prime} \}$) to wavenumbers $\tilde{\nu}=\sqrt{k_i^{\prime\prime}}/(2\pi \times 137 \times 5.29\times 10^{-9})$.  137 is the speed of light in atomic units and $5.29\times 10^{-9}$ converts bohr to cm.

More detail

Harmonic vibrational frequencies come from solving $F=ma$ assuming $E(x)=\frac{1}{2}k(x-x_0)^2 = \frac{1}{2}k^\prime l^2$ where $l=\sqrt{m}(x-x_0)$ and $k^\prime = k/m$. $F=-(\partial E/\partial x)$ so $F=ma$ becomes 
$$ \frac{\partial^2 l}{\partial t^2}+k^\prime l = 0 $$
The solution to this differential equation is 
$$ l = K \cos(\sqrt{k^\prime}t) +\varepsilon$$
where $K$ and $\varepsilon$ are arbitrary constants. 

The solution tells us that the displacement coordinate will increase and decrease in a repeating fashion every $\sqrt{k^\prime}/2\pi$ times, which is the definition of frequency ($\nu$). The units of $k^\prime$ are energy/(mass length$^2)$ or 1/time$^2$, because energy = (mass)(length/time)$^2$.  So the units of $\nu$ is 1/time.  This can be converted to a wavenumber ($\tilde{\nu}$, 1/length) by dividing $\nu$ by the speed of light.

Diatomic molecule 

You have 6 Cartesian coordinates $\{q_i\}$ and 6x6 Cartesian force constants (some have the same value),
$$ k_{ij} = \left( \frac{\partial^2 E}{\partial q_i \partial q_j} \right) $$
and it's not clear what to do with the cross terms.
So you diagonalize the matrix to get rid of the cross terms, but you need to mass-weight first for the reasons described above.
$$k^\prime_{ij} = \frac{1}{\sqrt{m_i m_j}}k_{ij}$$
After diagonalization of $\mathbf{k^\prime}$ you have 3N-5(6) non-zero force constants $\{ k_i^{\prime \prime}\}$ , which you can convert to wavenumbers by $\sqrt{k_i^{\prime \prime}}/2\pi c$