Navier-Stokes Convective term divergence of tensor is a row vector Consider the convective term from the Navier-Stokes equation (e.g. Navier–Stokes momentum equation (conservation form) from):
$$
\nabla \cdot (\rho \vec u \otimes \vec u ).
$$
Writing the outer product for three dimensional Euclidean space:
$$
\vec u \otimes \vec u = \rho
\begin{pmatrix}
  uu & vu & wu\\
  uv & vv & wv\\
  uw & vw & ww
\end{pmatrix}
$$
we will arrive at
\begin{align}
\nabla \cdot (\rho \vec u \otimes \vec u ) &=
\begin{pmatrix}
  \dfrac{\partial}{\partial x}\\
  \dfrac{\partial}{\partial y}\\
  \dfrac{\partial}{\partial z}
\end{pmatrix}
\cdot
\begin{pmatrix}
  \rho uu & \rho vu & \rho wu\\
  \rho uv & \rho vv & \rho wv\\
  \rho uw & \rho vw & \rho ww
\end{pmatrix} \\
&=
\begin{pmatrix}
  \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z}
\end{pmatrix}
\begin{pmatrix}
  \rho uu & \rho vu & \rho wu\\
  \rho uv & \rho vv & \rho wv\\
  \rho uw & \rho vw & \rho ww
\end{pmatrix} \\
& =
\begin{pmatrix}
  \dfrac{\partial \rho uu}{\partial x} 
+ \dfrac{\partial \rho uv}{\partial y} 
+ \dfrac{\partial \rho uw}{\partial z},
&   \dfrac{\partial \rho vu}{\partial x} 
+ \dfrac{\partial \rho vv}{\partial y} 
+ \dfrac{\partial \rho vw}{\partial z},
&   \dfrac{\partial \rho wu}{\partial x} 
+ \dfrac{\partial \rho wv}{\partial y} 
+ \dfrac{\partial \rho ww}{\partial z}
\end{pmatrix}
\end{align}
My confusion is that the result is a row vector (1 x 3) not a column one (3 x 1) which is the case for all other terms in the equation. Shouldn't the convective term be
$$
\left( \nabla \cdot (\rho \vec u \otimes \vec u ) \right)^T ~?
$$
If we consider the NS equations in convective form (see Navier–Stokes momentum equation (convective form) from the link above) the convective term:
$$
\vec u \cdot \nabla \vec u
$$
is in my opinion ambiguous. We can evaluate it like this $\vec u \cdot (\nabla \vec u)$ or like this $(\vec u \cdot \nabla) \vec u$. The former delivers a row vector and the latter a column vector. Some authors use explicitly the second form (see Incompressible Navier–Stokes equations (convective form) on the Wiki page)
Why is this term inconsistent (row vector) with other terms of NS equation? Doesn't it matter and I should ignore this fact? Do I miss something?
I found several questions about the convective term (here or here) but none of these addresses my issue.
 A: It should be written as $\rho ({\bf v}\cdot \nabla){\bf v}$, or in index notation as $\rho v_i \partial_i v_j$.
It's best to avoid the "row vector" and "column vector" language when working with tensor products. Indeed, in my opinion,  the use of the tensor product notation in the Wikipedia is unnecessarily confusing for the likely reader of the article.  I would write the momentum conservation equation as
$$
\partial_t(\rho v_i) +\partial_j (\rho v_i v_j +\delta_{ij}p)=\hbox{viscosity term}, 
$$
which, when combined with mass conservation
$$
\partial_t \rho+ \partial_i (\rho v_i)=0,
$$
leads to the Euler form of the NS equation
$$
\rho( \partial_t v_i +v_j \partial_j v_i)= - \partial_ip + \hbox{viscosity term}.
$$
A: 
My confusion is that the result is a row vector (1 x 3) not a column one (3 x 1) which is the case for all other terms in the equation.

My guess it's probably related to your converting a column vector to a row vector between the second equals sign. You should have something like,
$$\nabla\cdot\mathbf{u}\mathbf{u}=\left(\begin{array}{c}\partial_x \\ \partial_y \\ \partial_z \end{array}\right)\cdot\left(\begin{array}{ccc}u^2 & uv & uw \\ uv & v^2 & vw \\ uw & vw & w^2\end{array}\right)=\left(\begin{array}{c}(u^2)_x + (uv)_y + (uw)_z \\ (uv)_x + (v^2)_y + (vw)_z \\ (uw)_x + (vw)_y + (w^2)_z\end{array}\right)$$
which assuredly is a column vector.

...the convective term: $\vec{u}\cdot\nabla\vec{u}$  is in my opinion ambiguous. We can evaluate it like this $\vec{u}\cdot(\nabla\vec{u})$ or like this $(\vec{u}\cdot\nabla)\vec{u}$.

Fun fact: the two terms in the second sentence are actually identical, so it is not at all ambiguous. If you are careful with your math, it is easily provable; for simplicity, consider a 2D case:
$$ \left(\vec{u}\cdot\nabla\right)\vec u=\left(u\partial_x+v\partial_y\right)\left(\begin{array}{c}u \\ v\end{array}\right)=\left(\begin{array}{c}u\partial_xu+v\partial_yu \\ u\partial_xv+v\partial_yv\end{array}\right)=\left(\begin{array}{c}uu_x+vu_y \\ uv_x+vv_y\end{array}\right) $$
Now the other way,
$$ \vec{u}\cdot(\nabla\vec{u})=\left(\begin{array}{c}u \\ v\end{array}\right)\cdot\left(\begin{array}{cc}u_x & u_y \\ v_x & v_y\end{array}\right)=\left(\begin{array}{c}uu_x + vu_y \\ uv_x + vv_y\end{array}\right).
$$
