[Jackson uses the convention of a $\frac{1}{\sqrt{2\pi}}$ factor for each dimension, so there is a $\frac{1}{(2\pi)^2}$ in front of the integral in each direction.]
Start with the transform in the other direction, expressing $F(\vec k, \omega)$ in term of $F(\vec x, t)$:
$$ F(\vec k, \omega) = \int d^3x \int dt\ F(\vec x, t)\ e^{-i(\vec k \cdot \vec x - \omega t)} $$$$ F(\vec k, \omega) = \frac{1}{(2\pi)^2}\int d^3x \int dt\ F(\vec x, t)\ e^{-i(\vec k \cdot \vec x - \omega t)} $$
and substitute in the delta function for $F(\vec x, t)$:
$$ F(\vec k, \omega) = \int d^3x \int dt\ \delta(\vec x - \vec v t)\ e^{-i(\vec k \cdot \vec x - \omega t)} $$$$ F(\vec k, \omega) = \frac{1}{(2\pi)^2}\int d^3x \int dt\ \delta(\vec x - \vec v t)\ e^{-i(\vec k \cdot \vec x - \omega t)} $$
Doing the $d^3x$ integration and using the properties of the delta function just replaces the $\vec x$ with $\vec v t$ in the exponential, so we get:
$$ F(\vec k, \omega) = \int dt\ e^{-i(\vec k \cdot \vec v t - \omega t)} = \int dt\ e^{-i(\vec k \cdot \vec v - \omega) t} $$$$ F(\vec k, \omega) = \frac{1}{(2\pi)^2} \int dt\ e^{-i(\vec k \cdot \vec v t - \omega t)} = \frac{1}{(2\pi)^2} \int dt\ e^{-i(\vec k \cdot \vec v - \omega) t} $$ The last integral is recognized as a representation of the delta function: $$ F(\vec k, \omega) = \delta(\vec k \cdot \vec v - \omega) $$$$ F(\vec k, \omega) = \frac{1}{2\pi}\delta(\vec k \cdot \vec v - \omega) $$