One approach to work out the multi-variable Taylor expansion of $f(\vec{v})$ about a point $\vec{v}_0$ is to reduce it to the single variable case by writing the function
$ g(s) = f(\vec{v}_0 + s (\vec{v} - \vec{v}_0)) $
then
$ g(s) = \sum_{n=0}^{+\infty} \frac{s^n}{n!} g^{(n)}(0) $
The repeated derivatives of $g$ are repeated directional derivatives of $f$: (if this isn't clear, try working out $g'(s)$ yourself by whatever means you like)
$ g^{(n)}(s) = (\nabla_{\vec{v} - \vec{v}_0})^n f(\vec{v}_0 + s (\vec{v} - \vec{v}_0)) $
If you've only seen directional derivatives defined for unit vectors, the generalization to all vectors is the obvious one: $\nabla_{\vec{w}} h = \vec{w} \cdot \nabla h$
So the Taylor series is
$ f(\vec{v}_0 + s (\vec{v} - \vec{v}_0)) = \sum_{n=0}^{+\infty} \frac{s^n}{n!} \left( \nabla_{\vec{v} - \vec{v}_0}\right)^n f (\vec{v}_0) $
which simplifies to, if it converges,
$ \begin{align} f(\vec{v}) &= \sum_{n=0}^{+\infty} \frac{1}{n!} \left( \nabla_{\vec{v} - \vec{v}_0}\right)^n f (\vec{v}_0) \\ &= \sum_{n=0}^{+\infty} \frac{1}{n!} \left( (\vec{v} - \vec{v}_0) \cdot \nabla \right)^n f (\vec{v}_0) \end{align} $