$T(X,Y)=\frac{X+Y}{1-XY}$ is a power series which satisfies axiom of formal group.
My book reads this formal group comes from algebraic group $S: x^2+y^2=1$, with group law $*$:$(x_1,y_1)*(x_2,y_2)=(x_1x_2-y_1y_2,x_1y_2+x_2y_1)$.
$y/x$ is local parameter at $(1,0)$, thus expanding $*$ at origin gives formal group $T(X,Y)$.
What does 'expanding $*$ at origin gives formal group $T(X,Y)$' mean ?
I thought we can write $x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$, but calculating $xy-yx$ does not meet with $T(x,y)$. I'm sticking with how to gain tangent like formal group $T(X,Y)$ from group law of $S$ as my book reads.
The formal group law is visible as follows. Let $(x_1,y_1)*(x_2,y_2)=:(x_3,y_3)$, which is a multiplication of points near the origin $(1,0)$. We use the local parameter (in the sense below) $z:=y/x$, which is well-defined as long as $x\neq 0$, e.g., the point $(x,y)$ is near the origin. We denote by $z_i$ the value of the parameter corresponding to the points $(x_i,y_i)\ (i=1,2,3)$. Now "expand" $$\begin{split} z_3&=\frac{y_3}{x_3}=\frac{x_1y_2+x_2y_1}{x_1x_2-y_1y_2}\\ &=\frac{y_2/x_2+y_1/x_1}{1-(y_1/x_1)\cdot (y_2/x_2)}=\frac{z_1+z_2}{1-z_1z_2}, \end{split}$$ where the third equality is given by dividing both the numerator/denominator by $x_1x_2$. Thus we deduced the formal group law $z_3=T(z_1,z_2)$. This can be seen just the multiplication formula near the origin, expanded in terms of the local parameter $z$.
The terminology "local parameter" is frequently used to refer (any of) the generator of the maximal ideal of a DVR. In our case, this DVR is the local ring $\mathcal{O}_{G,e}$ at the identity $e$ in the algebraic group $G$, where $G=\mathrm{Spec}\ k[X,Y]/(X^2+Y^2-1)$ and $e=(1,0)(=(X-1,Y))$. For example, $z=y/x$ is a local parameter by the Jacobian criterion, namely $(\partial f/\partial x)(1,0)=2\neq 0$ (I believe you are in characteristic $\neq 2$) shows that $y$ is a local parameter, and the unit $1/x$ is multiplied so that our group law becomes simple. Of course we could have used another local parameter, in which case we get another group law and a (nontrivially looking) isomorphism between formal group laws.
EDIT: We shall prove that $y$ is a uniformizer of the DVR $\mathcal{O}_{G,e}$, where $G=\mathrm{Spec}\ k[x,y]/f,\ f=x^2+y^2-1$ and $e=(1,0)$. I think the proof of Jacobian criterion as in e.g. [Hartshorne's book Chapter I, Theorem 5.1] will be the simplest to apply. There it is proved that $m/m^2\simeq a/((f)+a^2)$, where $m$ is the maximal ideal of $\mathcal{O}_{G,e}$ and $a$ is the maximal ideal $(x-1,y)$ of $k[x,y]$. To see $y$ is a generator of the 1-dimensional $k$-linear space $m/m^2$, it is enough to show $y\not\in (f)+a^2$. But $$(f)+a^2=((x-1)(x+1)+y^2,(x-1)^2,(x-1)y,y^2)\subset (x-1,y^2),$$ so surely $y$ is not in this ideal. Note, here, that $a/a^2$ is a $k$-linear space with basis $x-1,y$ and in this space $$f=\frac{\partial f}{\partial x}(1,0)(x-1)+\frac{\partial f}{\partial y}(1,0)y\quad (\mathrm{mod}\ a^2)$$ holds by Taylor expansion. So, the condition $\frac{\partial f}{\partial x}(1,0)\neq 0$ really corresponds to $y$ being the basis of $a/((f)+a^2)$. This is why I tend to say "by Jacobian criterion...".