Is there a known simplification for the multi-index notation of $$ (\vec{v}\cdot \vec{A})^kf(\vec{x}) =\sum_{|\alpha|=k} \frac{k!}{\alpha!} (\vec{v} \circ \vec{A})^\alpha f(\vec{x}) = \sum_{|\alpha|=k} \frac{k!}{\alpha!}(\vec{v} \circ a(\vec{x})\vec{\nabla} )^\alpha f(\vec{x}) = ? $$ with the linear (vector) operator $ \vec{A} f(\vec{x}):=\left[a(\vec{x})\vec{\nabla} \right] f(\vec{x})$. Note that the scalar coefficient $a(\vec{x})$ has a spatial dependency. The goal is to find an expression with the multi-index notation of the nabla operator $\partial^\alpha$. For simplicity, the problem is restricted to $\alpha = (\alpha_1,\alpha_2)$. Thus the vectors and vector operators are two dimensional.
For the special case $a=const.$ we have $\sum_{|\alpha|=k} \frac{k!}{\alpha!}(\vec{v} \circ a(\vec{x})\vec{\nabla} )^\alpha f(\vec{x}) = \sum_{|\alpha|=k} \frac{k!}{\alpha!}\vec{v}^\alpha a^\alpha \partial^\alpha f(\vec{x})$.
Note that problem drops out from an power series expansion of the operator exponential $e^{\vec{v}\cdot \vec{A}} f(\vec{x})$.