The impression I got form the equation is that you model your system as being composed of a collection of point particles. The index $i$ indexes this set of point particles (as opposed to indexing the components of a vector). 

One could create a spatial velocity distribution by saying $\vec{v}(\vec{r})=\sum_i \vec{v}_i \delta(\vec{r} - \vec{r}_i)$, where $\vec{v}_i$ is the velocity of the $i$th particle and $\vec{r}_i$ is the position of the $i$th particle. Then the spatial fourier transform of this $\vec{v}(\vec{r})$ is $\vec{v}(\vec{k})$ or, as you have written it, $\vec{v}_\vec{k}$.

To get the dispersion relation, we put back in the time dependence of the velocities to get the function $\vec{v}_i(t)$. From this we form the velocity distribution $\vec{v}(\vec{r},t) = \sum_i \vec{v}_i(t) \delta(\vec{r} - \vec{r}_i)$. Now we can again do a spatial fourier transform to get $\vec{v}(\vec{k},t)$. We expect an oscillation to with wavevector $\vec{k}$ to osciallate at some frequency $\omega$, to find this frequency $\omega$, we take the temportal fourier transform of $\vec{v}(\vec{k},t)$ to get $\vec{v}(\vec{k},\omega)$. Now if we want to know the frequency corresponding to a given wavevector $\vec{k}_0$, then we just look at the frequency dependence of $\vec{v}(\vec{k}_0,\omega)$, and the $\omega$ for which this is non-zero give the frequency of the oscillation. So this procedure from getting an $\omega$ from a $k$ is a one way of getting the dispersion relation.

There is one more subtlety which is that $\vec{v}(\vec{k},\omega)$ has three components. Thus at a given $\vec{k}$ and $\omega$, the component of $\vec{v}(\vec{k},\omega)$ parallel to $\vec{k}$ may be non-zero but the perpendicular component may be zero. This tells you that longitudinal and transverse waves oscillate at different frequencies.