Suppose I want to compute $$ ({\mathbf{a}} \cdot {\mathbf{A}})\,\otimes\,({\mathbf{b}} \cdot {\mathbf{B}}) $$ where ${\mathbf{a}}$ and ${\mathbf{b}}$ are vectors, while $\mathbf A$ and $\mathbf B$ are vector operators. I know that $$ {\mathbf{A}} \cdot {\mathbf{B}} := {\mathrm{A}}_x {\mathrm{B}}_x + {\mathrm{A}}_y{\mathrm{B}}_y + {\mathrm{A}}_z{\mathrm{B}}_z $$ My question is how does "$\otimes$" distribute over $\,$"$\cdot$"$\,$?
A possible answer may be found by just expanding the two scalar products $$ \left({\mathrm{a}}_x {\mathrm{A}}_x + {\mathrm{a}}_y{\mathrm{A}}_y + {\mathrm{a}}_z{\mathrm{A}}_z\right) \otimes \left( {\mathrm{b}}_x {\mathrm{B}}_x + {\mathrm{b}}_y{\mathrm{B}}_y + {\mathrm{b}}_z{\mathrm{B}}_z\right) = \sum_{i=x,y,z} \mathrm{a}_i{\mathrm{A}}_i \otimes \sum_{j=x,y,z}{\mathrm{b}_j\mathrm{B}}_j $$
but I don't know if this is correct... are there really these many terms ($9$ in this case)?
Assuming each of $A$ and $B$ are operators on $V$ then each lives in $\text{Hom}(V,V)$ which is a vector space and hence your tensor product is a tensor of two such spaces: $$A \otimes B \in \text{Hom}(V,V) \otimes \text{Hom}(V,V)$$
And we know the algebraic rules of a tensor product of two spaces. If $\lambda$ is a scalar (remember that a vector space is equipped with scalar multiplication from elements of the field), then: $$(\lambda A +B) \otimes (C + D) = \lambda A \otimes C+ \lambda A \otimes D + B \otimes C + B \otimes D$$
$\boldsymbol{a}\cdot\boldsymbol{A}$ is just a shorthand for $a_1A_1 +a_2A_2+a_3A_3$ which is nothing but a sum of elements of $\text{Hom}(V,V)$ along with scalar multiplication and we know how $\otimes$ distributes over those (I stated that above). So yes, in short, what you wrote is correct and my post is supposed to give you justification of why this is right purely from the properties of a tensor product space.