$u(x,t)\cdot(\nabla\rho(x,t))$ versus $(u(x,t)\cdot\nabla)\rho(x,t)).$ I am new in the Navier Stokes Equation
In the Eulerian description of the fluid, if I have $g(t)=\rho(x(t),t)$, then:
$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+\frac{\partial \rho}{\partial x}(x(t),t)x'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+ u(x,t)\cdot(\nabla\rho(x,t)).$$
$u(x,t)=x'(t)$ is the velocity field of the flow of the fluid, $\rho:\Omega×[0,T] \rightarrow \mathbb{R}$ is the mass density, and $\nabla$ is the nabla operator.
MY QUESTION IS how did
$u(x,t)\cdot (\nabla\rho(x,t))$ become $(u(x,t)\cdot\nabla)\rho(x,t))?$
I searched on internet and they said that:

Generally the convective derivative of the field u·∇y, the one that contains the covariant derivative of the field, can be interpreted both as involving the streamline tensor derivative of the field u·(∇y), or as involving the streamline directional derivative of the field (u·∇) y, leading to the same result.

BUT I don't know why and how did this lead to the same result!!
 A: This is just because $\rho$ is a scalar function (i.e. takes values in $\mathbb R$). If you check
\begin{align} 
(\nabla \rho )_i &= \partial_i \rho,
\\
u\cdot\nabla &=\sum_i u_i \partial_i,
\end{align}
so
$$u\cdot (\nabla \rho) = \sum_i u_i \partial_i \rho = \sum_i(u_i \partial_i)\rho = (u\cdot\nabla) \rho.$$
The situation is slightly different if you were computing $g'$ for $g(t) = v(x(t),t)\in\mathbb R^n$. Then, you might wonder, what exactly is the meaning of
$$ u \cdot (\nabla v)?$$
In this case, $u$ is a vector, but $\nabla v$ is a matrix. As a matrix product, we have
$$ (u\cdot\nabla)v = (\nabla v)u.$$
The computation: writing components out in full (and using chain rule)
\begin{align} \frac d{dt} g_i'(t) 
&= \partial_t v_i(x,t) + \sum_j\partial_jv_i(x,t)\partial_t x_j  
\\&= \partial_t v_i(x,t) + \sum_j\partial_jv_i(x,t)u_j(x,t)  
\\&= \partial_t v_i(x,t) + \sum_ju_j(x,t)\partial_jv_i(x,t) \end{align}
In the last line you see exactly $\sum_ju_j(x,t)\partial_j =: u\cdot\nabla $ appear, and in the second last line, you see $\nabla v=( \partial_j v_i)_{i,j}$ multiplied on the right by the vector $u$. The two lines are equivalent because we wrote all the components out and at the level of scalars, everything commutes.
IMO $u\cdot(\nabla v)$ is dangerous notation in this case, it does not match the matrix product $u^T \nabla v$. Somehow it looks like you should be able to relate this with doing a "contraction" with the "left index" i.e. $a\cdot M := a_i M_{ij}$, but this is wrong because the Jacobian $(\nabla v)_{ij} = \delta_{\color{red} j} v_{\color{red} i}$ is the other way.
