Are these charts on the circle compatibly oriented? I've tried a few methods but I can't seem to work this one out.
Consider the charts
$$f(s) = (\cos s, \sin s) \in \mathbb{R}^2$$
for $-\pi < s < \pi$ and 
$$g(t)=(\frac{2t}{t^2 + 1}, \frac{t^2 - 1}{t^2 + 1})$$
for $t \in \mathbb{R}$. Are these charts on the circle compatibly oriented? 

The moving point $f(s)$ traces out the unit circle counterclockwise, starting and ending at $(-1,0)$, which, however is  excluded. Put $f\bigl(]{-\pi},\pi[\bigr)=:U$.
In order to better understand $g$ introduce an auxiliary variable $\tau$ that is bijectively and order preserving related to $t$ by
$$t=t(\tau)=\tan{\tau\over2}\qquad(-\pi<\tau<\pi)\ .$$
Then we can replace $g$ by its pullback
$$\eqalign{\tilde g(\tau):=g\bigl(t(\tau)\bigr)=g\left(\tan{\tau\over2}\right)=\left({2\tan{\tau\over2}\over\tan^2{\tau\over2}+1},{\tan^2{\tau\over2}-1\over \tan^2{\tau\over2}+1}\right)&=(\sin\tau, -\cos\tau)\cr &(-\pi<\tau<\pi)\ .\cr}$$
Wenn $\tau$ goes from $-\pi$ to $+\pi$ the point $\tilde g(\tau)$ traces out the unit circle counterclockwise as well, starting and ending at the point $(0,1)$ which is likewise  excluded. Put $g({\mathbb R})=\tilde g\bigl(]{-\pi},\pi[\bigr)=:V$.

It follows that the intersection $U\cap V$ of the two domains covered by the charts $f$ and $g$ is disconnected, which  is definitely undesirable. It implies that we have different transformation formulas between $s$ and $t$ in the two components of $U\cap V$.
For $-\pi<s<{\pi\over2}$ we have $\tau:=s+{\pi\over2}\in\ ]{-\pi},\pi[\ $ and therefore
$$f(s)=\bigl(\cos s,\sin s\bigr)=\bigl(\sin\tau,-\cos\tau)=\tilde g(\tau) =g\bigl(\tan{\tau\over2}\bigr)=g\Bigl(\tan\bigl({s\over2}+{\pi\over4}\bigr)\Bigr)\ . $$
So in this part of $U\cap V$ the parameter transformation reads $t=\tan\bigl({s\over2}+{\pi\over4}\bigr)$.
For ${\pi\over2}<s<\pi$ we have $\tau:=s-{3\pi\over2}\in\ ]{-\pi},\pi[\ $ and therefore
$$f(s)=\bigl(\cos s,\sin s\bigr)=\bigl(\sin\tau,-\cos\tau)=\tilde g(\tau) =g\bigl(\tan{\tau\over2}\bigr)=g\Bigl(\tan\bigl({s\over2}-{3\pi\over4}\bigr)\Bigr)\ . $$
So in this part of $U\cap V$ the parameter transformation reads $t=\tan\bigl({s\over2}-{3\pi\over4}\bigr)$.
As $\tan'(x)>0$ wherever defined the two charts are everywhere compatibly oriented.
