I found a problem which had $$\partial_i (A_i \vec{G})= (\vec{\nabla} .\vec{ A} )\vec{G}+ (\vec{A}.\nabla) \vec{G} $$ but my problem is what does $$\partial_i (A_i \vec{B})$$ even mean? it doesn't look like $(\vec{\nabla} .\vec{ A} )\vec{G} $ neither does it looks like $\nabla (\vec{A}.\vec{B})$ So what is it in terms of $\nabla $. Since the term $$\partial_i (A_i \vec{B})$$ needs to be integrated so it would be easier to write it in $\nabla$
Source: Problem 1.11 Modern Electromagantism by Zangwill
Let $C$ be an ordinary vector, and dot it with the dyad of the $A$ and $G$ vectors $$C\cdot(AG) = (C\cdot A)G$$ Similarly, the equation in your question can be written in dyadic notation as $$\nabla\cdot(AG) \;=\; (\nabla\cdot A)G \color{red}{+ A\cdot(\nabla G)}$$ Note that dot product only involves the $A$ vector, but differentiation occurs for both $A$ and $G$, and this gives rise to the $\rm\color{red}{second}$ term. The appearance of extra terms like this is what makes nabla notation tricky.
Also note that the dyadic product of two vectors is a second order tensor, i.e. $$M = AG \quad\iff\quad M_{ij} = A_iG_j$$
The quantity $(\nabla\cdot A)$ is the divergence of $A$ and is a scalar function.
The quantity $(\nabla\cdot M)$ is the divergence of $M$ and is a vector function.
The quantity $(\nabla G)$ is the gradient of $G;\,$ a ${\tt2}^{nd}$ order tensor function.