I just learned about the Kronecker product as a product of arrays. However, it seems that in all of the examples, when the Kronecker product is applied, the right side is multiplied with the components of the left side as scalar multiplication. Is this strictly necessary, or could the following be said?
$$ \begin{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} \\ \begin{bmatrix} 3 & 4 \end{bmatrix} \end{bmatrix} \otimes \begin{bmatrix} 5 \\ 6 \end{bmatrix} = \begin{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} \begin{bmatrix} 5 \\ 6 \end{bmatrix} \\ \begin{bmatrix} 3 & 4 \end{bmatrix} \begin{bmatrix} 5 \\ 6 \end{bmatrix} \end{bmatrix} = \begin{bmatrix} 17 \\ 39 \end{bmatrix} $$
$ \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\mm{\m{5&7\\6&8}} \def\c#1{\color{red}{#1}} \def\g#1{\color{blue}{#1}} $It's much more instructive to look at the product of two matrices, as displayed in the header of the Kronecker Product tagged page
Here is a Kronecker product related to your example $$\eqalign{ \m{1&2\\3&4}\otimes\mm = \m{1\mm&2\mm\\3\mm&4\mm} = \m{ \c{5} & 7 & 10 & 14 \\ 6 & 8 & \c{12} & 16 \\ \g{15} & 21 & 20 & 28 \\ 18 & 24 & \g{24} & 32 \\ } }$$ By contrast, your example shows a standard dot product, not a Kronecker product, i.e. it adds the two $\c{\rm red}$ numbers and the two $\g{\rm blue}$ numbers together and throws everything else away.