Is it enough to give a time-orientation to define a spin structure?

Maybe I got it wrong and my question doesn’t make sense, excuse me if that’s the case. For a smooth Lorentz 4-manifold $(M, g)$ with signature $(- + + +)$ is it enough to give a time-orientation to define a spin structure? Or is spatial orientation necessary too?

More precisely, let $pi_{O}: F(M) rightarrow M$ be the orthonormal frame bundle (a principal $O(1, 3)$-bundle). So, if I choose the set $L$ consisting only of the orthonormal bases such that $e_{0}$ is a timelike vector ($g(e_{0}, e_{0}) < 0$), future-pointing (in the class designated as future) and $e_{1}, e_{2}, e_{3}$ are spacelike vectors, then do I have a principal $SO(1, 3)$-bundle $(L, SO(1, 3), pi_{SO})$?

$$SO(1, 3) = {A in O(1, 3): det(A) = 1 and A_{0}^{0} geq 1},$$
where $Ae_{0} = (A_{0}^{0}, A_{1}^{1}, A_{2}^{2}, A_{3}^{3})^{T}$. And the spin structure comes from the double covering $Lambda: SL(2, mathbb{C}) rightarrow SO(1, 3)$.

My doubt is because the time-orientation is not a common orientation for manifolds (in the sense of the Jacobian sign), as there are examples of non-orientable manifolds that are time-orientable. Appreciate.