The set of orthonormal bases of $\mathbb R^n$ is isomorphic to the orthogonal group $O(n)$, which has dimension $\frac{n(n-1)}2$. (Just view the vectors in the basis as the columns of a matrix.) Locally, the orthonormal frame bundle $\mathcal OM$ of a Riemannian $n$-manifold $(M,g)$ thus is isomorphic to the product of an open subset in $\mathbb R^n$ with $O(n)$, hence it has dimension $n+\frac{n(n-1)}2=\frac{n(n+1)}2$.

The relation to the isometry group comes from the fact that any isometry $f$ of $M$ lifts to a diffeomorphism $F$ of $\mathcal OM$. Basically, this is given by $F(p,e_1,\dots,e_n):=(f(p),T_pf(e_1),\dots,T_pf(e_n))$ (note that the derivative $T_pf$ is an orthogonal map, thus sending an orthonormal basis to an orthonormal basis). If $M$ is connected, then one can argue directly, that if $f$ and $g$ are isometries such that $f(p)=g(p)$ and $T_pf=T_pg$, then $f=g$. (Using that an isometry sends geodesics to geodesics, it follows that $f$ and $g$ coincide in a neighborhood of $p$ and then $f=g$ follows by connectedness.) This shows that fixing a point $(p,e_1,\dots,e_n)\in\mathcal OM$, the map $f\mapsto F(p,e_1,\dots,e_n)$ defines an injective map from the isometry group to $\mathcal OM$. It needs a bit more effort (for example using that the Levi-Civita connection provides a trivialization of the tangent bundle $T(\mathcal OM)$) to see that the isometry group is a Lie group and that the map constructed above is an embedding.