From the Wikipedia entry on the derivation of the Navier-Stokes equations, the continuity equations for scalar quantities has the following form： $$ \frac{\partial \varphi}{\partial t} + \nabla \cdot(\varphi \vec u) + \vec s = \vec 0\label{1}\tag{1} $$ Since $$ \nabla \cdot (\vec a \otimes \vec b) = (\nabla \cdot \vec a) \vec b + \vec a \cdot \nabla \vec b\label{2}\tag{2} $$ and $$ \nabla \cdot (s \vec v) = \nabla s \cdot \vec v + s \nabla \cdot \vec v \label{3}\tag{3} $$ I get $$ \nabla \cdot ( \vec a \otimes \vec b) = (\nabla \cdot (\vec a b_x), \nabla \cdot (\vec a b_y))\label{4}\tag{4} $$ When $\varphi$ is a vector, its components respectively satisfy \eqref{1}, that is, $$ \begin{cases} \dfrac{\partial \varphi_x}{\partial t} + \nabla \cdot(\varphi_x \vec u) + s_x = 0 \\ \dfrac{\partial \varphi_y}{\partial t} + \nabla \cdot(\varphi_y \vec u) + s_y = 0 \end{cases}\label{5}\tag{5} $$
According to \eqref{4}, the two equations \eqref{5} above could be written as the single one below: $$ \frac{\partial \vec \varphi}{\partial t} + \nabla \cdot(\vec u \otimes \vec \varphi) + \vec s = \vec 0 $$ But in the Wikipedia entry, $\vec \varphi$ is at the left to $\vec u$, i.e. $$ \frac{\partial \vec \varphi}{\partial t} + \nabla \cdot(\vec \varphi \otimes \vec u) + \vec s = \vec 0 $$ This implies that after substituting $\vec \varphi$ by $\rho \vec u$, my version of the equation can’t be simplified to the elegant form $$ \rho\left(\frac{\partial \vec u}{\partial t}+\vec u \cdot \nabla \vec u\right) = \vec s $$ which is basically Newton’s second law.
That implies my version was wrong. Where did I go wrong?