Let $M$ be an oriented smooth surface, and let \begin{equation}\mathbb{Z_2} \to \operatorname{Spin}(2) \overset{\theta}{\to} \operatorname{SO}(2) \end{equation} be the usual central extension defining the $\operatorname{Spin}$ group $\operatorname{Spin}(2)$ as the connected double cover of $\operatorname{SO}(2)$.
The usual definition of a spin structure on $M$ is
Definition 1: A spin structure on $M$ is principal $\operatorname{Spin}(2)$ bundle $\xi: P_{\operatorname{Spin}(2)} \to M$ together with a double covering $\rho: P_{\operatorname{Spin}(2)} \to P_{\operatorname{SO}(2)}$ ($P_{\operatorname{SO}(2)}$ being the oriented orthonormal frame bundle) that is equivariant, i.e. $\rho(pg) = \rho(p)\theta(g)$ $\forall p \in P_{\operatorname{Spin}(2)}, g \in \operatorname{Spin}(2)$ (equivalently, the covering map restricts to the connected double cover on each fiber).
Taking classifying spaces of the central extension one obtains a fibration sequence $$B\mathbb{Z}_2 \to B \operatorname{Spin}(2) \overset{B\theta}{\to} B \operatorname{SO}(2)$$
which leads to the "tangential structure" definition of a spin structure:
Definition 2 Let $T: M \to B\operatorname{SO}(2)$ be the classifying map of the orientation structure on $M$. A spin structure on $M$ is a lift $\phi$ of $T$ to $B\operatorname{Spin}(2)$ via $B\theta$.
Why are the two definitions equivalent? To get definition (1) from definition (2), note that the maps $M \to B\operatorname{SO}(2)$ and $M \to B\operatorname{Spin}(2)$ give us bundles $P_{\operatorname{Spin}(2)}$ and $P_{\operatorname{SO}(2)}$ over $M$. We have the following diagram: 
$P_{\operatorname{Spin}(2)}$ and $P_{\operatorname{SO}(2)}$ come from pulling back along $\phi$ and $T$ respectively. $B\theta$ induces the bundle $E\operatorname{Spin}(2) \times_{\operatorname{Spin}(2),\theta} \operatorname{SO}(2)$ in the following way: $\operatorname{SO}(2)$ can be turned into a left $\operatorname{Spin}(2)$-space by $g.h$ = $\theta(g)h$ where $g \in \operatorname{Spin}(2)$ and $\operatorname{SO}(2)$. Then $E\operatorname{Spin}(2) \times_{\operatorname{Spin}(2),\theta} \operatorname{SO}(2)$ is just the usual balanced product.
There is an obvious map $E \operatorname{Spin}(2) \to E\operatorname{Spin}(2) \times_{\operatorname{Spin}(2),\theta} \operatorname{SO}(2)$ which makes everything commute (I think). Define the red arrow as the composition $E \operatorname{Spin}(2) \to E\operatorname{Spin}(2) \times_{\operatorname{Spin}(2),\theta} \operatorname{SO}(2) \to E \operatorname{SO}(2)$. Via some diagram chasing, we get that the composites $P_{\operatorname{Spin}(2)} \to E\operatorname{Spin}(2) \to E \operatorname{SO}(2) \to B\operatorname{Spin}(2)$ and $P_{\operatorname{Spin}(2)} \to M \to B \operatorname{SO}(2)$ are equal. By the universal property of the pullback, we finally get the red dotted arrow $P_{\operatorname{Spin}(2)} \to P_{\operatorname{SO}(2)}$.
Why is the dotted red arrow a double cover?