The material derivative in vector form is, $$\frac{DQ}{Dt}=\frac{\partial Q}{\partial t}+(V\cdot\nabla) Q$$ Where $Q$ is the fluid property. I didn't understand the following thing from here
The form only works for Cartesian coordinates because it not invariant under coordinate transformation. This means that it does not hold true when using curvilinear coordinates such as Cylindrical or Spherical. Fortunately, we can write it using invariant form as follows, $$ \begin{align} (V\cdot\nabla) Q &=\frac12[ \nabla(V\cdot Q)\\ & -V\times(\nabla\times Q)-Q\times\nabla\times V\\ &- \nabla\times(V\times Q) + V(\nabla\cdot Q)-Q(\nabla\cdot V) ] \end{align} \qquad (1) $$
I couldn't understand from where the invariant form come? Is there any proof of that? I got some intuition when the $Q$ is velocity field from here.
The blog post you cited obtains the material derivative as $$ \frac{DQ}{Dt} = \frac{\partial Q}{\partial t} + V\cdot\nabla Q $$ where $Q$ is a scalar function. This is invariant under change of coordinates. Now, if you replace the scalar function $Q$ with a vector field $A$, the equation you get is $$ \frac{DA}{Dt} = \frac{\partial A}{\partial t} + (V\cdot\nabla) A $$ I guess the post says that this form is not invariant under change of coordinates because some people can interpret $(V\cdot\nabla) A$ incorrectly as $\sum_i V^i \partial_i A^j$. This last expression is, indeed, not invariant under change of coordinates.
If $A$ is a vector field, the proper interpretation of $(V\cdot\nabla) A$ is the directional derivative of $A$ along $V$, which is another vector field whose components are $$((V\cdot\nabla) A)^j = \sum_i V^i\nabla_iA^j$$ where $\nabla_iA^j$ are the components of the covariant derivative of $A$, and not the gradients of the components of $A$, which are the $\partial_iA^j$.
I guess the author of the post wanted to avoid mentioning the covariant derivative, which requires a bit of background in differential geometry. In that case you can express $(V\cdot\nabla) A$ purely in terms of vector operators as in the blog post: $$ \def\dt{\!\cdot\!} \def\cr{\!\times\!} (V\!\cdot\!\nabla) A = \frac12[ \nabla(V\dt A) -V\cr(\nabla\cr A)-A\cr(\nabla\cr V) - \nabla\cr(V\cr A) + (\nabla\dt A)V-(\nabla\dt V)A ]. $$ You can find this identity on the wiki page about vector calculus identities. As to where this expression comes from, if you are familiar with tensor notation you can get from the right hand side to the left one using the translations $$ \begin{align} (\nabla f)^i &= \nabla^i f \\ (\nabla \times B)^i &= E^{ijk}\nabla_iB_j \\ (B\times C)^i &= E^{ijk}B_jC_k \\ \nabla\cdot B &= \nabla_iB^i \end{align} $$ where a repeated index up and down means summation and $E^{ijk}$ are the Levi-Civita tensors. Or, if you are into graphical notations, like myself, you can use these translations into Penrose's graphical notation.