Your question is a bit ambiguous, but here's how I choose to interpret it. Choose any local orthonormal moving frame $e_1,\dots,e_n$ on $S^n$. Indeed, if $e_0$ is the position vector, the matrix formed by $e_0,e_1,\dots,e_n$ gives us a mapping to $O(n+1)$. Thinking of $e_A$ ($A=0,\dots,n$) as maps to $\Bbb R^{n+1}$, we write $de_A = \sum\limits_{B=0}^n \omega_{AB}e_B$, we'll have $\omega_{0j} = \omega_j$ ($j=1,\dots,n$) the dual coframe to the original moving frame and $(\omega_{ij})$ will be the connection matrix of $1$-forms.

Now $d(de_A) = 0$ gives us $d\omega_{AB} = \sum\limits_C \omega_{AC}\wedge\omega_{CB}$. In particular, the entries of the curvature matrix of $2$-forms are
$$\Omega_{ij} = d\omega_{ij} - \sum_{k=1}^n\omega_{ik}\wedge\omega_{kj} =
\sum_{C=0}^n \omega_{iC}\wedge\omega_{Cj} - \sum_{k=1}^n\omega_{ik}\wedge\omega_{kj} = \omega_{i0}\wedge\omega_{0j} = -\omega_i\wedge\omega_j.$$
This tells us, in particular, that every sectional curvature is $1$.

(There are always sign issues with the structure equations depending on whether we think of $e_A$ as column vectors or row vectors, i.e., depending on whether we have $O(n+1)$ acting on the left or on the right on the orthonormal frame bundle. I'll leave you to sort this out ...)