How meaningful on an unique $(c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3)$?? Since the few (smooth) vector fields can denote themself in the unit vectors,
$$\boldsymbol{\dbinom{B}{C}}=\dbinom{b_1\boldsymbol i+b_2\boldsymbol j+b_3\boldsymbol k}{c_1\boldsymbol i+c_2\boldsymbol j+c_3\boldsymbol k}$$
we have
$$(c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3)+(b_1\nabla c_1+b_2\nabla c_2+b_3\nabla c_3)\\=((\nabla b_1)c_1+b_1\nabla c_1)+((\nabla b_2)c_2+b_2\nabla c_2)+((\nabla b_3)c_3+b_3\nabla c_3)\\=\nabla(b_1c_1)+\nabla(b_2c_2)+\nabla(b_3c_3)\\=\nabla(b_1c_1+b_2c_2+b_3c_3)\\=\boldsymbol\nabla(\boldsymbol B\bullet\boldsymbol C)$$
I v found it once I was expanding
$$\begin{cases}\boldsymbol{(\nabla\times B)\times C}&\equiv(\boldsymbol C\bullet\boldsymbol\nabla)\boldsymbol B-(c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3)\\\boldsymbol{(B\times\nabla)\times C}&\equiv(b_1\nabla c_1+b_2\nabla c_2+b_3\nabla c_3)-\boldsymbol B(\boldsymbol\nabla\bullet\boldsymbol C)\end{cases}$$ seperately, then we have
$$\boldsymbol{(\nabla\times B)\times\boldsymbol C}-\boldsymbol{(B\times\nabla)\times C}+\boldsymbol\nabla(\boldsymbol B\bullet\boldsymbol C)=(\boldsymbol C\bullet\boldsymbol\nabla)\boldsymbol B+\boldsymbol B(\boldsymbol\nabla\bullet\boldsymbol C)\ .\ :)$$
However I think an unique $(c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3)$ is sticky and v no ideas to his meanings and collapsing in the vector fields. Who can tutor me?? Ur all answers r welcome!! :D
 A: The identity you have is correct:$\newcommand{\b}{\boldsymbol}$
$$
\nabla (\b{B}\cdot \b{C}) = (\b{B}\cdot \nabla) \b{C} + (\b{C}\cdot \nabla) \b{B} - (\nabla \times\b{B})\times\b{C} - (\nabla \times\b{C})\times\b{B} \tag{$\dagger$}
$$
I rewrite it a bit to make it easier to understand and memorize.

$(c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3)$ has no specific meaning, but is sometimes denoted as the following in physics:
$$
c_1\nabla b_1+c_2\nabla b_2+c_3\nabla b_3 := \nabla_{\b{B}}(\b{B}\cdot \b{C})
$$
this means $\b{B}$ is differentiated while $\b{C}$ is held constant. It is defined as:
$$
\nabla_{\b{B}}(\b{B}\cdot \b{C}) = (\b{C}\cdot \nabla) \b{B} - (\nabla \times\b{B})\times\b{C} \tag{1}
$$
which is what you got up there. 
Both terms in the right hand side of (1) have a geometric meaning: $(\b{C}\cdot \nabla) \b{B} $ is the directional derivative of $\b{B}$ in the direction of $\b{C}$, scaled by the magnitude of $\b{C}$; $(\nabla \times\b{B})\times\b{C}$ is the curl of $\b{B}$ projected to the plane which has $\b{C}$ as normal vector, then rotated a right angle such that it is perpendicular to both $\nabla \times \b{B}$ and $\b{C}$, also scaled by the magnitude of $\b{C}$.
Coincidentally, both terms' geometric meaning relies on a fixed $\b{C}$. We fixed a $\b{C}$ at a point, then we are literally decomposing the rate of change of the other vector $\b{B}$  into two parts, one part is the gradient along the direction of $\b{C}$, the other is the gradient perpendicular to $\b{C}$.
Similarly:
$$
b_1\nabla c_1+b_2\nabla c_2+b_3\nabla c_3 := \nabla_{\b{C}}(\b{B}\cdot \b{C}) = (\b{B}\cdot \nabla) \b{C} - (\nabla \times\b{C})\times\b{B} \tag{2}
$$

Lastly, instead of your way of obtaining the identity, we can directly consider the following:
$$
\nabla (\b{B}\cdot \b{C}) = \nabla_{\b{B}}(\b{B}\cdot \b{C}) + \nabla_{\b{C}}(\b{B}\cdot \b{C})
$$
which is the product rule for vector, then together with (1) and (2), we have the identity $(\dagger)$.
