Coordinates of the tensor of the second fundamental form for a submanifold I'm self-study Riemannian Geometry in order to be able to understand this lecture notes about Mean Curvature Flow. I'm reading the first chapter, which is a review of Riemannian Geometry, and I'm stuck in the following part:

For local choice of unit normal vector field $\nu$, the second fundamental form of $M$, a $(0,2)-$tensor is given by
$$A(V,W) = - \langle S_V W, \nu \rangle_e = \langle W, \overline{\nabla}_V \nu \rangle_e$$
or in coordinates $A = (h_{ij})$ by
$$h_{ij} = - \left\langle \frac{\partial^2 F}{\partial x_jx_i},\nu \right\rangle_e = \left\langle \frac{\partial F}{\partial x_i},\frac{\partial \nu}{\partial x_j} \right\rangle_e.$$

A relevant information on the lecture notes is

We restrict ourselves to manifolds of codimension $1$ in an Euclidean ambient
  space, i.e. we consider a $n$-dimensional smooth manifold $M$, without boundary,
  either closed or complete and non-compact and an immersion (or embedding)
$$F: M \longrightarrow \mathbb{R}^{n+1}$$
We call the image $F(M)$ a hypersurface. We will often identify points on M with their image under the immersion, if there is no risk of confusion. Let $x = (x_1, \cdots , x_n)$ be a local coordinate system on M[...] We denote by
$$g_{ij} = \left\langle \frac{\partial F}{\partial x_i}, \frac{\partial F}{\partial x_j}  \right\rangle$$

My doubt is why the expression for $h_{ij}$ is that from above? 
$\textbf{My attempt in order to understand the coordinates $h_{ij}$:}$
Firstly, I think that $h_{ij} = A \left( \frac{\partial F}{\partial x_i}, \frac{\partial F}{\partial x_j} \right)$, then
$h_{ij} = - \left\langle S_{\frac{\partial F}{\partial x_i}} \frac{\partial F}{\partial x_j}, \nu \right\rangle_e = - \left\langle \left( \overline{\nabla}_{\frac{\partial F}{\partial x_i}} \frac{\partial F}{\partial x_j} \right)^{\perp}, \nu \right\rangle_e$
and
$h_{ij} = \left\langle \frac{\partial F}{\partial x_j}, \overline{\nabla}_{\frac{\partial F}{\partial x_i}} \nu \right\rangle_e$
Secondly, I think that if $U_{\alpha} \subset \mathbb{R}^n$ are open sets and $\phi_{\alpha}: U_{\alpha} \longrightarrow M$ are charts of the manifold $M$, then $F \circ \phi_{\alpha}: U_{\alpha} \longrightarrow F(M) \subset \mathbb{R}^{n+1}$ are charts of the submanifold $F(M)$, therefore the $\frac{\partial}{\partial x_i} = d (F \circ \phi_{\alpha}) (e_i)$ for $i = 1, \cdots, n$, where $e_i$ is an element of canonic basis of $\mathbb{R}^n$, but I don't sure about this because in the literature of Riemmanian Geometry $\frac{\partial}{\partial x_i}$ denote an element of coordinate basis of $T_pM$.
Thanks in advance!
 A: Your attempt is mostly correct with a remark in order. In differential geometry the notation plays a crucial role, and the devil is in the details. What is happening  here is a lot of tacit or implicit identifications, that are mentioned but not really reflected in the notation.
First of all, there is an identification along the charts $\phi_a$, taking into account that they are diffemorphisms. Thus, the tangential vector fields on $M$ are generated by 
$$
\frac{\partial F}{\partial x^i} := dF(\frac{\partial}{\partial x^i})
$$
Secondly, there is an identification along the pullback $F^*$, and the fields $\frac{\partial F}{\partial x^i}$ are thought as the same as the fields from the coordinate frame $\frac{\partial}{\partial x^i}$.
For instance, the first fundamental form, aka the metric $g$ on $M$ is given on page 3 of your notes as
$$
g_{i j} := \langle \frac{\partial F}{\partial x^i}, \frac{\partial F}{\partial x^j} \rangle
$$
that in the pullback can be written as familiar as $g_{i j} = g(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})$, which is actually a definition of the intrinsic metric as the pullback of the ambient metric.
Next, let's take a closer look at $h_{i j} = A(\frac{\partial F}{\partial x^i}, \frac{\partial F}{\partial x^j})$, following the notation and definition, adopted in the notes (and the question):
$$
\begin{align}
h_{i j} & = A(\frac{\partial F}{\partial x^i}, \frac{\partial F}{\partial x^j}) = - \langle \big( S_{\frac{\partial F}{\partial x^i}} \frac{\partial F}{\partial x^j} \big)^{\perp}, \nu \rangle_{e}
= - \langle S_{\frac{\partial F}{\partial x^i}} \frac{\partial F}{\partial x^j} , \nu \rangle_{e} \\
& = - \langle \overline{\nabla}_{\frac{\partial F}{\partial x^i}} \frac{\partial F}{\partial x^j} , \nu \rangle_{e} = - \langle \overline{\nabla}_{dF(\frac{\partial}{\partial x^i})} \frac{\partial F}{\partial x^j} , \nu \rangle_{e} \equiv - \langle \overline{\nabla}_{\frac{\partial}{\partial x^i}} \frac{\partial F}{\partial x^j}, \nu \rangle_{e} \\
& = - \langle \frac{\partial^2 F}{\partial x^i \partial x^j}, \nu \rangle_{e}
\end{align} 
$$
where the last line follows from the definition of $\overline{\nabla}$:
$$
\overline{\nabla}_{\frac{\partial}{\partial x^i}} Y = \big( \frac{\partial Y^{1}}{\partial x^i}, \dots, \frac{\partial Y^{n+1}}{\partial x^i}  \big)
$$
and
$$
\frac{\partial F}{\partial x^i} = 
\begin{pmatrix}
\frac{\partial F^{1}}{\partial x^i} \\
\dots \\
\frac{\partial F^{n+1}}{\partial x^i}
\end{pmatrix}
$$
and you need to spot the cheating with pullbacks :)
