In my Riemannian geometry course there is a great emphasis on being able to compute the coefficients of the second fundamental form of some surface in $\mathbb{R}^4$ given by an immersion. Let me give a concrete example.
Consider the function $f:\mathbb{R}\times (0, 2\pi)\to \mathbb{R}^4$, $f(u, v)=(u\cos v, u\sin v, u^3\cos v, u^3\sin v)$. Then $N=f(\mathbb{R}\times (0, 2\pi))$, i.e. the image of $f$, is a surface in $\mathbb{R}^4$ (a submanifold with dimension $2$). The coefficients of the metric on $N$ are given by $$g_{11}=\left\langle \frac{\partial f}{\partial u}, \frac{\partial f}{\partial u}\right\rangle,\quad g_{12}=g_{21}=\left\langle \frac{\partial f}{\partial u}, \frac{\partial f}{\partial v}\right\rangle, \quad g_{22}=\left\langle \frac{\partial f}{\partial v}, \frac{\partial f}{\partial v}\right\rangle,$$ that is we work with the metric induced by $\mathbb{R}^4$ on this submanifold.
When we compute the coefficients of the second fundamental form, we have to compute, for instance, $$\left\langle h(e_1, e_2), e_3\right\rangle,$$ where $\{e_1, e_2\}$ is an orthonormal basis in $T_p M$ and $\{e_3, e_4\}$ is an orthonormal basis in $T^{\perp}_p M$.
Finding these bases is straightforward in this case, we just take $e_1=\frac{1}{\sqrt{g_{11}}}\frac{\partial f}{\partial u}$ and $e_2=\frac{1}{\sqrt{g_{22}}}\frac{\partial f}{\partial v}$ and then apply the Gram-Schmidt procedure.
What I don't understand is why in this context $$\left\langle h(e_1, e_2), e_3\right\rangle=\frac{1}{\sqrt{g_{11}g_{22}}}\left \langle\frac{\partial^2 f}{\partial u \partial v}, e_3\right\rangle.$$ This boils down to $$h\left(\frac{\partial f}{\partial u}, \frac{\partial f}{\partial v}\right)=\frac{\partial^2 f}{\partial u \partial v},$$ which is essentially what causes me trouble.
In my course we defined the second fundamental form in general for a submanifold of a Riemannian manifold. I am not sure if there are multiple ways to do it, but just to be precise we seem to have done it in the way Do Carmo does in Chapter 6 of his Riemannian geometry book. I don't understand how to get from that abstract definition to the result I need in this concrete example.
EDIT: I will write down $e_3$ and $e_4$ by the method I was taught. We denote by $J_0:\mathbb{R}^4\to \mathbb{R}^4$, $J_0(x, y, z, w)=(-z, -w, x, y)$ the standard complex structure on $\mathbb{R}^4$. It is straightforward to see that $$\left\langle J_0 e_1, e_2\right\rangle=0$$ and this tells us that $N$ is a Lagrangian surface in $\mathbb{R}^4$, so we may take $e_3=J_0e_1$ and $e_4=J_0e_2$ (this is what I referred to as Gram-Schmidt above; I am not entirely sure how this relates to the classical Gram-Schmidt procedure, but this is what we were told to do without too many explanations).
To make my post easier to follow, I now include the explicit formulas I got for the coeffcients of the metric and for the orthonormal frame: $$g_{11}=1+9u^4, \quad g_{12}=g_{21}=0, \quad g_{22}=u^2(1+u^4)$$ $$e_1=\frac{1}{\sqrt{1+9u^4}}(\cos v, \sin v, 3u^2\cos v, 3u^2 \sin v)$$ $$e_2=\frac{1}{\sqrt{1+u^4}}(-\sin v, \cos v, -u^2 \sin v, u^2\cos v)$$ $$e_3=\frac{1}{\sqrt{1+9u^4}}(-3u^2\cos v, -3u^2\sin v, \cos v, \sin v)$$ $$e_4=\frac{1}{\sqrt{1+u^4}}(u^2\sin v, -u^2\cos v, -\sin v, \cos v).$$
$\newcommand\R{\mathbb{R}}$Let $O \subset \R^2$ be open and let $f: O \rightarrow \R^4$ be an embedding. The first fundamental form is $$ g = g_{ij}\,dx^i\,dx^j, $$ where $$ g_{ij} = \partial_if\cdot\partial_jf. $$ For each $x \in O$, there is an orthogonal decomposition $$ \R^4 = T_x \oplus N_x, $$ where $T_x$ is the subspace spanned by $\partial_1f(x), \partial_2f(x)$ and $N_x$ is the $2$-plane orthogonal to $T_x$. Let $$ \pi_x^\perp: \R^4 \rightarrow N_x $$ denote orthogonal projection onto $N_x$.
The second fundamental form can now be defined as follows (or inferred from a different but equivalent definition): Given any tangent vectors $$ v= v^1\partial_1f + v^2\partial_2f, w = w^1\partial_1f + w^2\partial_2f, $$ the second fundamental form is defined to bve $$ h(v,w) = \pi_x^\perp(v^iw^j\partial^2_{ij}f) = v^iw^j\pi_x^\perp(\partial^2_{ij}f) \in N_x. $$ In other words, $$ h = h_{ij}\,dx^i\,dx^j, $$ where $$ h_{ij} = \pi^\perp_x(\partial^2_{ij}f) \in N_x. $$ Strictly speaking, $h \in S^2T^*_x\otimes N_x$.
In particular, if $(e_1,e_2,e_3,e_4)$ is an adapted orthonormal frame, then for any $1 \le i,j \le 2$, \begin{align*} e_1\cdot h_{ij} &= 0\\ e_2\cdot h_{ij} &= 0\\ e_3\cdot h_{ij} &= e_3\cdot\partial^2_{ij}f\\ e_4\cdot h_{ij} &= e_4\cdot\partial^2_{ij}f\\ \end{align*}
By the way, the tangential component of $\partial^2_{ij}f$ is essentially the Christoffel symbols of $g$ with respect to the coordinates $(x^1,x^2) \in O$.