$\DeclareMathOperator{\Spin}{Spin}$Let $X$ be an oriented smooth $n$-manifold with the frame bundle $\pi_{SO} \colon F_{SO} \to X$. Then the bundle of spin frames is a $\Spin(n)$-bundle $\pi_{\Spin} \colon F_{\Spin} \to X$ together with a $2$-fold equivariant covering $\pi \colon F_{\Spin} \to F_{SO}$ such that $\pi_{\Spin} = \pi_{SO} \circ \pi$.
A bundle of spin frames can be considered as a group reduction of the bundle $\pi_{SO} \colon F_{SO} \to X$ along the covering homomorphism $\Spin(n) \to SO(n)$, which is, by definition, a bundle $\pi_{\Spin} \colon F_{\Spin} \to X$ together with an isomorphism of $SO(n)$-bundles $F_{\Spin} \times_{\Spin(n)} SO(n) \to F_{SO}$. My question is whether the second definition also implies the first one in the sense that any $\Spin(n)$-bundle associated to $F_{SO}$ via the covering projection $\Spin(n) \to SO(n)$ will define a spin structure (a bundle of spin frames)?
The terminology you use ("group reduction ... along the covering homomorphism", "$\text{Spin}(n)$-bundle associated to $F_{SO}$ via the covering projection") is a bit misleading. It is not a reduction but an extension of structure group, and this is not a functorial process.
Anyway, the answer to your question is yes, if you include in your first definition that $\pi$ is equivariant with respect to the covering homomorphism. Given an equivariant $2$-fold covering $\pi:F_\text{Spin}\to F_{SO}$, the map $F_\text{Spin}\times SO(n)\to F_{SO}$ defined by $(u,A)\mapsto \pi(u)\cdot A$ induces an isomorphism $F_\text{Spin}\times_{\text{Spin}(n)}SO(n)\to F_{SO}$. Conversely, given an isomorphism $F_\text{Spin}\times_{\text{Spin}(n)}SO(n)\to F_{SO}$, you just compose with the obvious inclusion $F_\text{Spin}\to F_\text{Spin}\times_{\text{Spin}(n)}SO(n)$ (mapping $u$ to the class of $(u,id)$) to obtain an equivariant projection $\pi$.