I would like to show that fixing the orientation of $ k $ -manifold smooth connected $ S $ in $ \mathbb {R} ^ n $ is equivalent to fixing a frame for one of its tangent spaces. (Source: Zorich, Mathematical Analysis II, ‘orientation of surfaces’, pag.175)
What I know is that different orientations correspond to orienting atlases containing maps that cannot be consistent with maps in other orienting atlases of other equivalence classes, when their domains of action overlap.
How could we pass from this fact, to the fact that it is enough to fix a frame anywhere $ x_0 \in S $ to determine the orientation of $ S $?
It seems from the question that you are defining an orientation of a connected, orientable manifold $M$ to be a maximal oriented atlas (i.e. a maximal atlas such that all transition functions have Jacobians with uniformly positive determinant). If you are instead using an equivalence class of atlases, the maximal atlas is just the union of the equivalence class.
One can show $M$ has exactly two such orientations, which we can call $\mathcal{A}$ and $-\mathcal{A}$, and any overlapping charts $(U,\varphi)\in\mathcal{A}$ and $(V,\psi)\in-\mathcal{A}$ are incompatible, i.e the Jacobian of the transition function between them has uniformly negative determinant.
Choose a basis $E=(e_1,\dots,e_n)$ for $T_pM$. Given a chart $(U,\varphi)$ with $p\in U$, we define the coordinate representation of $E$, denoted $E^i_j$,as the matrix whose columns are coordinate representations of the $e_j$, i.e. $E^i_j=dx^i(e_j)$. We say that $E$ is consistently oriented with $\varphi$ if $E^i_j$ has positive determinant. When changing to a different chart $\bar{\varphi}$, we have $\bar{E}^i_j=\frac{\partial \bar{x}^i}{\partial x^k}E^k_j$, and so $\det(\bar{E}^i_j)=\det(d\Phi)\det(E^i_j)$, where $\Phi$ is the transition function.
Given an orientation $\mathcal{A}$, we say $E$ is consistently oriented with $\mathcal{A}$ if $E$ is consitently oriented with some (and thus any) chart around $p$ in $\mathcal{A}$. Note that $E$ is consistently oriented with exactly one of the two orientations of $M$.
Therefore, given any basis for any tangent space, we may choose the unique orientation consistently oriented with that basis.