Second fundamental form and embeddings Let $\Sigma$ be a smooth hypersurface of a $d$ dimensional smooth Riemannian manifold $(\mathcal M, G)$;
we may see $G_x$ as a mapping from $T_x(\mathcal M)$ into $T_x^*(\mathcal M)$ so that 
$$
\langle G_x X, X\rangle_{T^*_x(\mathcal M), T_x(\mathcal M)}=G_x(X).
$$
 Working in a neighborhood of a point of $\Sigma$ we may consider a parametrization from a chart $U$ of $\Sigma$ into a chart $W$ of $\mathcal M$,
$$
\mathbb R^{d-1}\supset U\ni u\mapsto x(u)\in W\subset\mathbb R^d, \quad \text{rank $x'(u)=d-1.$}
$$
We define the first fundamental form $g_u$ as 
$$
\langle g_u T, T\rangle_{E^*, E}=\langle G_{x(u)} x'(u)T, x'(u)T\rangle_{F^*, F}, \quad T\in \mathbb R^{d-1}=E, \quad F=T_{x(u)}(\mathcal M),
$$
and $g_u$ is obviously a positive definite quadratic form on $E$. We introduce $N(u)$ as a normal vector to $\Sigma$ at $x(u)$ and that vector $N(u)$ belongs to $F$ and is such that 
$$
\langle G_{x(u)} x'(u)T, N(u)\rangle_{F^*, F}\equiv 0.
$$
We define the covector $\nu(u)=G_{x(u)} N(u)$ and we have
$
\langle \nu(u),x'(u)T\rangle_{F^*, F}\equiv 0,
$
so that 
$$
\langle \nu'(u) T,x'(u)T\rangle_{F^*, F}+\langle \nu(u),x''(u)T^2\rangle_{F^*, F}=0.
$$
I want to define the second fundamental form in the direction $N$  as 
$$
\langle\omega_u T, T\rangle_{E^*, E}=\langle \nu(u),x''(u)T^2\rangle_{F^*, F},
$$
and the Gauss curvature as 
$$
K=\frac{\det(\omega_u)}{\det (g_u)}.
$$
The above description is standard when $\mathcal M$ is the Euclidean $\mathbb R^d$.
Is it correct in this more general case ? 
 A: To be honest, I find your formulas hard to follow. I hope the following is useful:
I define the second fundamental form for a submanifold $\Sigma \subset \mathcal{M}$ of any codimension as follows: Let $X \in T_x\Sigma$ and $Y$ be a tangent vector field along $T_*\Sigma$. There are two different connections, the Levi-Civita connection $\nabla^\Sigma$ induced by the metric on $\Sigma$ and the connection $\nabla^{\mathcal{M}}$ induced by the metric on $\mathcal{M}$. It is straightforward to check that $\nabla^{\mathcal{M}}_XY$ is well defined, even though $Y$ is defined only on $\Sigma$. Then $\nabla^{\mathcal{M}}_XY$ can be decomposed into two components, one tangent to $\Sigma$ and the other normal. It turns out that the tangential component is $\nabla^{\Sigma}_XY$ and the normal component is the second fundamental form (viewed as a normal-vector-valued symmetric tensor on $T_*\Sigma$), which I'll denote by $H$:
$$
\nabla^{\mathcal{M}}_XY = \nabla^{\Sigma}_XY + H(X,Y).
$$
It is also straightforward to show that $H(X,Y) = H(Y,X)$. Since the definition of $H$ shows that $H$ depends only on the value of $X$ at the point $x \in \Sigma$, the symmetry of $H$ implies it is a tensor.
In codimension $1$, the standard second fundamental is simply $\nu\cdot H$, where $\nu$ is a unit normal vector$. It is also easy to see that the definition above is the same as the standard one using the Gauss map.
Also, as Oliver Nash points out, the second fundamental form can be viewed as a bundle map $H: S^2T_*\Sigma \rightarrow N_*\Sigma$ (this is the adjoint of Oliver's ap) or, equivalently, a section of the bundle $N_*\otimes S^2T^*$.
