For a two-dimensional smooth manifold $M$ with a Riemannian metric, we can consider the unit tangent bundle $S\to M$, which is a circle bundle over $M$. I am trying to show that if $S\to M$ is a "principal" circle bundle, then $M$ must be orientable. Here is my try: Suppose $S\to M$ is a principal circle bundle. Then there is a right action of $S^1$ on $S$. Then we can define orientation on each tangent space of $M$ by declaring $\{v, v\cdot i\}$ to be a positive basis for any nonzero tangent vector $v$. This implies that $M$ is orientable.
Does my argument work? I should show that the orientation on tangent spaces are well-defined (does not depend on the choice of $v$), but I got stuck showing this. Any hints?