I don't really get this identity : $$ \frac{\partial\Phi}{\partial n} = \vec n . \nabla \Phi $$ So gradient is: $$ \nabla \Phi = \frac{\partial\Phi}{\partial x} \vec i +\frac{\partial\Phi}{\partial y} \vec j +\frac{\partial\Phi}{\partial z} \vec k $$ Normal vector is:
$$ \vec n = n_x \vec i + n_y \vec j+ n_z \vec k $$ RHS of identity:
$$\vec n . \nabla \Phi = n_x \frac{\partial\Phi}{\partial x} + n_y \frac{\partial\Phi}{\partial y} + n_z \frac{\partial\Phi}{\partial z} $$
So must be:
$$ \frac{\partial\Phi}{\partial n} = n_x \frac{\partial\Phi}{\partial x} + n_y \frac{\partial\Phi}{\partial y} + n_z \frac{\partial\Phi}{\partial z} $$
Now is my question, what is $n_x $ , so the components of the normal vector
Can this be true: $$n_x = \frac{\partial x}{\partial n} $$
The inner product $\mathbf{A}\cdot \mathbf{\hat{B}}$ tells you how much of the vector $\mathbf{A}$ lies along the direction of the unitary vector $\mathbf{\hat{B}}$. If it is zero it means that they are perpendicular, if you get back $|\mathbf{A}|$ it means they were parallel.
Now, the gradient $\nabla \Phi$ is a vector and $\mathbf{\hat{n}}$ is usually an unitary vector. The product
$$ \mathbf{\hat{n}} \cdot \nabla \Phi $$
Is then how much of the vector $\nabla\Phi$ is projected in the direction of $\mathbf{\hat{n}}$, or say in another words, it tells you what is rate of change of $\Phi$ along $\mathbf{\hat{n}}$, hence the name: directional derivative, usually denoted as
$$ \frac{\partial \Phi}{\partial \mathbf{\hat{n}}} = \mathbf{\hat{n}} \cdot \nabla \Phi $$
or as $\nabla_{\mathbf{\hat{u}}}\Phi$. But just to emphasize, it does not mean you are taking the derivative with respect a vector, only means that you are calculating the projection of the derivative along said vector