In the book Topology, Geometry and Gauge Fields Interactions by Gregory L. Naber, he defines the standard orientation of $S^n$ (viewed as a subset of $\mathbb{R}^{n+1}$), $n\geq 2$ by the oriented atlas given by the following two charts
\begin{equation} \varphi_S \colon S^n -\{N\} \longrightarrow \mathbb{R}^n, \quad \varphi_S(x) = \left(\frac{x^1}{1-x^{n+1}}, \ldots, \frac{x^{n}}{1-x^{n+1}}\right) \end{equation}
and
\begin{equation} \tilde{\varphi}_N \colon S^n -\{S\} \longrightarrow \mathbb{R}^n, \quad \tilde{\varphi}_N(x) = \left(-\frac{x^1}{1+x^{n+1}}, \ldots, \frac{x^{n}}{1+x^{n+1}}\right) \end{equation}
where $\varphi_S$ is just the usual stereographic projection map. Now, the tangent space for any $p\in S^n$ can be thought of as a subspace of $T_p\mathbb{R}^{n+1}$ by identifying any $v \in T_pS^n$ with $\iota_{\ast p}v$ where $\iota\colon S^n \hookrightarrow \mathbb{R}^{n+1}$, and any tangent vector in $T_p\mathbb{R}^{n+1}$ can be identified with a vector in $\mathbb{R}^{n+1}$ by the canonical isomorphism $v_1\left.\frac{\partial}{\partial x^1}\right\lvert_p + \cdots + v_{n+1}\left.\frac{\partial}{\partial x^{n+1}}\right\lvert_p \mapsto (v_1, \ldots, v_{n+1})$. Therefore, to reduce notational clutter, I will denote elements of $T_pS^n$ just as vectors in $\mathbb{R}^{n+1}$ under this identification.
The claim that is made in the book is that for any $p\in S^n$, $e_1,\ldots, e_n \in T_pS^n$, then if $\{e_1, \ldots e_n\}$ is in the standard orientation on $S^n$, then $\{p, e_1, \ldots, e_n\}$ is oriented as a vector of $\mathbb{R}^{n+1}$.
Here is my attempt thus far: Let $x^1, \ldots, x^n$ be the coordinate functions of $\varphi_S$, then $(\left.\frac{\partial}{\partial x^1}\right\rvert_p, \ldots, \left.\frac{\partial}{\partial x^n}\right\rvert_p)$ is in the orientation of $S^n$ for each $p \in S^n-\{N\}$, so my thought is that it would be nice to establish that $\{p, \left.\frac{\partial}{\partial x^1}\right\rvert_p, \ldots, \left.\frac{\partial}{\partial x^n}\right\rvert_p\}$ is oriented as a vector of $\mathbb{R}^{n+1}$. However, I've run into the issue where this is only the case if $n$ is odd. Here are my computations:
We can compute the image of $\left.\frac{\partial}{\partial x^1}\right\rvert_p$ under the canonical isomorphism by considering the curve $\alpha(t) = \varphi_S^{-1}(\varphi_S(p) + te_1)$ and finding $\alpha'(0)$.
Now
\begin{equation} \varphi_S^{-1}(y) = \frac{1}{1+\lVert y\rVert^2}(2y^1, \ldots, 2y^n, \lVert y \rVert^2 - 1), \end{equation}
therefore,
\begin{equation} \alpha(t) = \frac{1}{1+\lVert \beta(t)\rVert^2}\left(\frac{2p^1}{1-p^{n+1}}+2t, \frac{2p^2}{1-p^{n+1}}, \ldots, \frac{2p^n}{1-p^{n+1}}, \lVert \beta(t)\rVert^2-1\right), \end{equation}
where $\beta(t) = \varphi_S(p) + te_1$. Moreover, $\lVert \beta(t) \rVert^2 = t^2 + 2t\langle \varphi_S(p), e_1\rangle + \lVert \varphi_S(p)\rVert^2$ so $\frac{d}{dt}\lVert \beta(t) \rVert^2 = 2t + \frac{2p^1}{1-p^{n+1}}$. Then
\begin{align} \frac{d}{dt} \alpha(t) = &-\frac{1}{(1+\lVert \beta(t)\rVert^2)^2}\left( 2t + \frac{2p^1}{1-p^{n+1}}\right) \left(\frac{2p^1}{1-p^{n+1}}+2t, \frac{2p^2}{1-p^{n+1}}, \ldots, \frac{2p^n}{1-p^{n+1}}, \lVert \beta(t)\rVert^2-1\right) \\ &+\frac{1}{1+\lVert\beta(t)\rVert^2}\left(2, 0, \ldots, 0, 2t + \frac{2p^1}{1-p^{n+1}}\right), \end{align}
so
\begin{align} \alpha'(0) = -\frac{1}{1+\lVert \varphi_S(p)\rVert^2}\left(\frac{2p^1}{1-p^{n+1}}\right)p + \frac{1}{1+\lVert \varphi_S(p)\rVert^2}(2e_1 + \frac{2p^1}{1-p^{n+1}}e_{n+1}). \end{align}
Now, notice that $\langle \alpha'(0),p \rangle = \frac{1}{1+\lVert \varphi_S(p)\rVert^2}\left(-\frac{2p^1}{1-p^{n+1}} +p^1+\frac{2p^1}{1-p^{n+1}}p^{n+1}\right) = \frac{1}{1+\lVert \varphi_S(p)\rVert^2}\left(\frac{-2p^1+p^1-p^1p^{n+1}+p^1p^{n+1}}{1-p^{n+1}} \right) = 0$, which is what we expect. So I'm fairly certain that this expression is indeed correct. Now, in the same way, we find
\begin{equation} \left.\frac{\partial}{\partial x^i}\right\rvert_p = -\frac{1}{1+\lVert \varphi_S(p)\rVert^2}\left(\frac{2p^i}{1-p^{n+1}}\right)p + \frac{1}{1+\lVert \varphi_S(p)\rVert^2}(2e_i + \frac{2p^i}{1-p^{n+1}}e_{n+1}), \end{equation}
so we are left with evaluating $\det(p, \left.\frac{\partial}{\partial x^1}\right\rvert_p, \ldots, \left.\frac{\partial}{\partial x^n}\right\rvert_p)$. Now, since $p$ is one of the vectors, we can add $\frac{1}{1+\lVert \varphi_S(p)\rVert^2}\left(\frac{2p^i}{1-p^{n+1}}\right)p$ to the corresponding $\left.\frac{\partial}{\partial x^i}\right\rvert_p$, so we have
\begin{equation} \det\left(p, \left.\frac{\partial}{\partial x^1}\right\rvert_p, \ldots, \left.\frac{\partial}{\partial x^n}\right\rvert_p\right) = \left( \frac{1}{1+\lVert \varphi_S(p)\rVert^2}\right)^n\det\begin{pmatrix}p^1 & 2 & 0 & \cdots & 0 \\ p^2 & 0 & 2 & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ p^{n+1} & \frac{2p^1}{1-p^{n+1}} & \frac{2p^2}{1-p^{n+1}}& \cdots & \frac{2p^n}{1-p^{n+1}}\end{pmatrix}. \end{equation}
Now, concentrating on the matrix in the determinant, we can get rid of the $p^i$ in the first column by subtracting from it the $p_i/2 \times $ the $(i+1)^\text{th}$ column. This then leaves the $n1$-entry of the matrix to be $p^{n+1}-\frac{(p^1)^2 + \cdots + (p^n)^2}{1-p^{n+1}} = p^{n+1}-\frac{1-(p^{n+1})^2}{1-p^{n+1}} = p^{n+1} - (1+p^{n+1}) = -1$. So the resulting expression becomes
\begin{equation} \left( \frac{1}{1+\lVert \varphi_S(p)\rVert^2}\right)^n\det\begin{pmatrix}0 & 2 & 0 & \cdots & 0 \\ 0 & 0 & 2 & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ -1 & \frac{2p^1}{1-p^{n+1}} & \frac{2p^2}{1-p^{n+1}}& \cdots & \frac{2p^n}{1-p^{n+1}}\end{pmatrix}, \end{equation}
which is positive if $n+1$ is even and negative if $n+1$ is odd.
Just a tldr; based on my computations, $\varphi_S$ is orientation preserving whenever $n$ is odd and orientation reversing whenever $n$ is even, which differs from the claim by Naber. In particular, if we think of the 2-sphere, we can easily visualise the tangent vectors. If we look at the point $(1,0,0)$, then moving in the x-y plane in the positive $x$ direction, then the image of this point under $\varphi_S^{-1}$ seems to go upwards, i.e. $\frac{\partial}{\partial x^1}_{(1,0,0)}\propto (0,0,1)$. On the other hand, if we move in the positive $y$ direction, the image of this point under $\varphi_S^{-1}$ follows the direction of motion, i.e. $\frac{\partial}{\partial x^2}_{(1,0,0)}\propto (0,1,0)$. Then heuristically, the matrix we get by adjoining $p$ to the front something like
\begin{equation} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & b \\ 0 & a & 0 \end{pmatrix}, \end{equation}
which obviously has negative determinant. So my question is: Is the claim by Naber correct, or is my argument flawed?