Let us assume that $E$ is an oriented Riemannian vector bundle equipped with spin structure. Therefore there is a $spin(n)$ principial bundle $spin(E)$ and an equivariant map $\eta:spin(E) \to SO(E)$ where $SO(E)$ is $SO(n)$ bundle of oriented orthonormal frames. In the book ,,Spin Geometry" by Lawson and Michelsohn it is claimed that the map $\eta$ corresponds (on fibers) to the usual $2:1$ covering $\rho:spin(n) \to SO(n)$. So I understand it as follows: for principial bundles $spin(E)$ and $SO(n)$ there are equivariant isomorphisms $\psi:spin(E)|_x \to spin(n)$ and $\varphi:SO(E)|_x \to SO(n)$ and we have to prove that $$\varphi \circ \eta=\rho \circ \psi$$ Since both sides are equivariant is suffices to show it for one element, say for $\psi^{-1}(e) \in spin(E)|_x$ where $e \in spin(n)$ is the identity.
How to show this equality?
Let us first clarify the definition of spin structure.
In the language of principal bundles, a spin structure on $E$ is defined as a $\rho$-reduction of $SO(E)$. This means, by definition, that a spin structure on $E$ is a tupel $(Spin(E),\eta$), where
(Note that we do not require $\eta$ to be a double covering. However, it follows from the definition of a spin structure, that $\eta$ is in fact a double covering. Roughly speaking, $\rho$ is a double covering and the local properties of $\rho$ carry on to $\eta$, so $\eta$ is also a double covering.)
In the following I try to answer your question. We choose a basepoint $p$ and and $x\in Spin(E)_p$. We have isomorphisms
$\psi\colon Spin(n)\to Spin(E)_p$, $g\mapsto x\cdot g$,
$\varphi\colon SO(n)\to SO(E)_p $, $A\mapsto \eta(x)\cdot A$.
$\psi$ and $\varphi$ are isomorphisms, since the action of $Spin(n)$ and $SO(n)$ is simply transtive on $Spin(E)_p$ and $SO(E)_p$, respectively.
We have that
$\eta(\psi(g))=\eta(x\cdot g)=\eta(x)\cdot \rho(g)=\varphi(\rho(g))$, hence,
$\eta\circ\psi=\varphi\circ \rho$.