Let $M$ be an orientable manifold, with chosen base point $q$ and chosen Riemannian metric. The extension of the question arises from the principal bundle $SO(n)\to F(M)\to M$. 

There is another bundle giving rise to the same extension, namely $\mathbb{RP}^\infty\to \tilde{M}\to M$, and there are three equivalent ways of constructing it. On the one hand, it can be obtained from stabilization to $SO(\infty)$ plus change of fibre along the action $SO(\infty)\times\mathbb{RP}^\infty\to\mathbb{RP}^\infty$. On the other hand, it can be obtained by passing to the stable tangent bundle and then taking the associated $\mathbb{RP}^\infty$-bundle. Third, the $\mathbb{RP}^\infty$-bundle is the pullback of the universal $\mathbb{RP}^\infty$-bundle $\mathbb{RP}^\infty\to\{\operatorname{pt}\}\to K(\mathbb{Z}/2,2)$ along the classifying map $M\to BSO(n)\to K(\mathbb{Z}/2,2)$. Under the identification $[M,K(\mathbb{Z}/2,2)]\cong H^2(M,\mathbb{Z}/2)$, this is the Stiefel-Whitney class $w_2$. (Note that the classifying map of the frame bundle can be identified with the classifying map for the tangent bundle, the frame bundle arises from choice of structure group reduction from $GL_n(\mathbb{R})$ to $SO(n)$ applied to the tangent bundle. This choice of reduction of structure group is the same as a choice of a Riemannian metric.)

I think, obstruction theory shows that a splitting of the extension is the same as a section of the bundle $\mathbb{RP}^\infty\to \tilde{M}\to M$, because $\mathbb{RP}^\infty\cong K(\mathbb{Z}/2,1)$. This way, one could see that the extension splits if and only if $w_2=0$.

In the general case where $M$ is not necessarily orientable, then $\pi_1^{\operatorname{or}}(M)$ is the fundamental group of the orientation cover, the pullback of $BSO(n)\to BO(n)$ along the classifying map $M\to BO(n)$. I think the above arguments can be applied to the orientation cover and then characterize the splitting of the extension as triviality of $w_2$ for the orientation cover.