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 $S^1$.

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 $D^2$. The (oriented) unit tangent vector to $S^1$ at the point $(\cos(0),\sin(0)) = (1,0)$ is just $e_y$ in the frame used to trivialize $SO(D^2)$ above, and at any other point $(\cos \theta, \sin \theta)$ 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 $S^1$.