Confusion on the orientability of the unit circle $\mathbb{S}^{1}$. I'd like to check that $\mathbb{S}^{1}$ is orientable. In order to do this, we check the jacobian matrix of the change of coordinates has positive determinant. 
Since there are two choices of atlases, I checked both of them, which brought me confusions. 
For the first charts, $\mathcal{A}=\{(U,\varphi), (V,\psi)\}$ where $$U:=\{(\cos\alpha,\sin\alpha):\alpha\in (0,2\pi)\}=\mathbb{S}^{1}\setminus\{(1,0)\},\ \varphi:U\longrightarrow\mathbb{R},\ \varphi(\cos\alpha,\sin\alpha)=\alpha,\ \alpha\in(0,2\pi)$$ $$V:=\{(\cos\alpha,\sin\alpha):\alpha\in (-\pi, \pi)\}=\mathbb{S}^{1}\setminus\{(-1,0)\},\ \psi:V\longrightarrow\mathbb{R},\ \psi(\cos\alpha,\sin\alpha)=\alpha,\ \alpha\in(-\pi, \pi).$$ 
things are fine, since on $U\cap V=\mathbb{S}^{1}\setminus\{(1,0), (-1,0)\}$, we have the change of coordinate $$\psi\circ\varphi^{-1}:\varphi(U\cap V)\longrightarrow U\cap V\longrightarrow\psi(U\cap V)$$ $$\alpha\mapsto (\cos\alpha, \sin\alpha)\mapsto
 \left\{
  \begin{array}{ll}
  \alpha\ \text{if}\ \alpha\in (0,\pi)\\
  \alpha-2\pi\ \text{if}\ \alpha\in (\pi,2\pi).
  \end{array}
 \right.$$
so that the jacobian of the change of coordinate is always the $1\times 1$ matrix $(1)$ and thus has determinant $1$, thus $\mathbb{S}^{1}$ is orientable. 
However, the confusion is from the second collection of charts,  namely the collection $\mathcal{A}:=\{(U_{1}, \varphi_{1}), (U_{2}, \varphi_{2}),(U_{2}, \varphi_{2}),(U_{2}, \varphi_{2})\},$ where $$U_{1}:\{(x,y)\in\mathbb{S}^{1}:x>0\},\ \varphi_{1}:U_{1}\longrightarrow\mathbb{R},\ \varphi_{1}(x,y)=y,$$ $$U_{2}:\{(x,y)\in\mathbb{S}^{1}:y>0\},\ \varphi_{2}:U_{2}\longrightarrow\mathbb{R},\ \varphi_{2}(x,y)=x,$$ $$U_{3}:\{(x,y)\in\mathbb{S}^{1}:x<0\},\ \varphi_{3}:U_{3}\longrightarrow\mathbb{R},\ \varphi_{3}(x,y)=y,$$ $$U_{4}:\{(x,y)\in\mathbb{S}^{1}:y<0\},\ \varphi_{4}:U_{4}\longrightarrow\mathbb{R},\ \varphi_{4}(x,y)=x.$$
Then,  on $U_{1}\cap U_{2}=\{x,y\in\mathbb{R}: x,y\in (0,1)\}$, we have  the change of coordinate  $$\varphi_{1}\circ \varphi_{2}^{-1}:\varphi_{2}(U_{1}\cap U_{2})=(0,1)\longrightarrow U_{1}\cap U_{2}\longrightarrow\varphi_{1}(U_{1}\cap U_{2})=(0,1)$$ $$x\mapsto (x,\sqrt{1-x^{2}})\mapsto \sqrt{1-x^{2}},$$ so that the jacobian is a $1\times 1$ matrix $\Big(-\dfrac{1}{\sqrt{1-x^{2}}}\Big)$.
Then, note that the determinant is negative. 
What happens between these two atlases? Did I make any mistakes in my computation?
Thank you!
 A: Okay, as both Angina Seng and augustoC pointed out, I mixed up the definition. 
Note that the definition of orientability only requires you to find ONE atlas such that the Jacobian matrix of the change of coordinate has positive determinant. It does NOT require every atlas of a manifold has positive determinant Jacobian matrix. 
Hence, since the first atlas satisfies the requirement,  $\mathbb{S}^{1}$ is without doubt orientable, and the second atlas not working does not bring us a contradiction. 
As also pointed out by augustoC, the difference between these two atlases lies in the counterclockwise and clockwise moving. The second atlas, for instance, $\varphi_{2}$ has the opposite orientation of $\varphi_{1}$ over the part of $\mathbb{S}^{1}$ in the first quadrant. Note that as we move over $(0,1)$ from small to larger values, $\varphi_{2}$ travels the quarter counterclockwise while $\varphi_{1}$ does it clockwise. Hence, we need to "flip" the orientation of $\varphi_{2}$ by orienting its codomain with the opposite orientation. 
Since neither of them wanted to post an answer, I am just answering my own question to close this post. (I just don't like it remaining open forever....)
