As Fabian pointed out in the comments, you have to be more careful about how you trivialize $SO(D^2)$. I'm going to use the standard coordinates $(x,y)$ on $\mathbb{R}^2$ (note that these are not global coordinates on $D^2$, but they still trivialize the frame bundle). Thus we have a global section of $SO(D^2)$ which assigns to each point in $D^2$ the standard orthonormal basis $(e_x,e_y)$, and the action of $S^1 = SO(2)$ on $SO(D^2) \cong D^2 \times S^1$ is by counterclockwise rotation of this basis. It is fairly clear from this picture that the principal spin bundle is $Spin(D^2) \cong D^2 \times S^1$, and the double cover $Spin(D^2) \to SO(D^2)$ is the identity on $D^2$ and the doubling map on $S^1$.
Now let's figure out how $SO(S^1) \cong S^1$ sits inside $SO(D^2)$ using this trivialization. The boundary circle is the set of points $(\cos \theta, \sin \theta)$ in $D^2$, so let's use the angular coordinate $\theta$ to describe points on the circle (remembering that $\theta = 0$ is identified with $\theta = 2\pi$). D^2$. The (oriented) unit tangent vector to$S^1$at the point corresponding to$\theta (\cos(0),\sin(0)) = 0$(1,0)$ is just $e_y$ in the frame used to trivialize $SO(D^2)$ above, and at any other point $\theta$ (\cos \theta, \sin \theta)$it is just$R_\theta e_y$where$R_\theta$denotes counterclockwise rotation by the angle$\theta$. So if we denote the oriented unit vector tangent to$\theta \in S^1$by$e_\theta$then the embedding$SO(S^1) \to SO(D^2) \cong D^2 \times S^1$is given by$e_\theta \mapsto ((\cos \theta, \sin \theta), \theta)$. Finally, in this picture it is clear that the inverse image of the set$\lbrace ((\cos \theta, \sin \theta), \theta) \rbrace$under the map$Spin(D^2) \to SO(D^2)$is the connected double cover of$S^1$. 1 As Fabian pointed out in the comments, you have to be more careful about how you trivialize$SO(D^2)$. I'm going to use the standard coordinates$(x,y)$on$\mathbb{R}^2$(note that these are not global coordinates on$D^2$, but they still trivialize the frame bundle). Thus we have a global section of$SO(D^2)$which assigns to each point in$D^2$the standard orthonormal basis$(e_x,e_y)$, and the action of$S^1 = SO(2)$on$SO(D^2) \cong D^2 \times S^1$is by counterclockwise rotation of this basis. It is fairly clear from this picture that the principal spin bundle is$Spin(D^2) \cong D^2 \times S^1$, and the double cover$Spin(D^2) \to SO(D^2)$is the identity on$D^2$and the doubling map on$S^1$. Now let's figure out how$SO(S^1) \cong S^1$sits inside$SO(D^2)$using this trivialization. The boundary circle is the set of points$(\cos \theta, \sin \theta)$in$D^2$, so let's use the angular coordinate$\theta$to describe points on the circle (remembering that$\theta = 0$is identified with$\theta = 2\pi$). The (oriented) unit tangent vector to$S^1$at the point corresponding to$\theta = 0$is just$e_y$in the frame used to trivialize$SO(D^2)$above, and at any other point$\theta$it is just$R_\theta e_y$where$R_\theta$denotes counterclockwise rotation by the angle$\theta$. So if we denote the oriented unit vector tangent to$\theta \in S^1$by$e_\theta$then the embedding$SO(S^1) \to SO(D^2) \cong D^2 \times S^1$is given by$e_\theta \mapsto ((\cos \theta, \sin \theta), \theta)$. Finally, in this picture it is clear that the inverse image of the set$\lbrace ((\cos \theta, \sin \theta), \theta) \rbrace$under the map$Spin(D^2) \to SO(D^2)$is the connected double cover of$S^1\$.