The components of the vector field are $X_1(x, y) = -y$ and $X_2(x, y) = x$. These components define a system of ODEs for a curve:
$$ \dot{\gamma}(t) = X(\gamma(t)), $$
or more explicitely
$$
\begin{align}
\dot{\gamma_1}(t) &= X_1(\gamma(t)) = -\gamma_2(t), \\
\dot{\gamma_2}(t) &= X_2(\gamma(t)) = \gamma_1(t).
\end{align}$$
The solution $\gamma(t)$ is the flow generated by the vector field. The function $\sigma$ solves the given system of ODES, therefore it is a flow. For further information take a look at this script from the ETH Zurich
A chart of $S^1$ can be $\phi^{-1}$ where
$\phi: (0,2\pi)\to S^1$
$\phi(\theta):=(cos(\theta),\sin(\theta))$
Now you say that
$i_*(\partial_\theta)=a\partial_x+b\partial_y$
where $(x,y)$ are the coordinates of $\mathbb{R}^2$
We want calculate $a,b\in \mathbb{R}$ but
$a=i_*(\partial_\theta)(x)=\partial_\theta(x\circ (i\circ (\phi^{-1})^{-1}))=\partial_\theta(cos(\theta))=-sin(\theta)$
while
$b=i_*(\partial_\theta)(y)=\partial_\theta(y\circ (\phi^{-1})^{-1}))=\partial_\theta(sin(\theta))=cos(\theta)$
So
$i_*(\partial_\theta)=-sin(\theta)\partial_x+cos(\theta)\partial_y$
You can observe that $i_*$ is an injective linear map because there is not $\theta$ such that $-sin(\theta)=cos(\theta)=0$ so $i$ is an immersion map.
It is not correct to write $i^*:T^*S^1\to T^*\mathbb{R}^2$ because the differentital of a smooth map is a contro-variant functor, so you must have
$i^*:T^*\mathbb{R}^2\to T^*S^1$.
Now
$i^*(dx)=\alpha d\theta$
and $i^*(dy)=\beta d\theta$
for some $\alpha,\beta$, so
$\alpha=i^*(dx)(\partial_\theta)=dx(i_*(\partial_\theta))=dx(a\partial_x+b\partial_y)=a=-sin(\theta)$ while
$\beta= i^*(dy)(\partial_\theta)=dy(i_*(\partial_\theta))=dx(a\partial_x+b\partial_y)=b=cos(\theta)$
Then
$i^*(dx)=-sin(\theta) d\theta$
and $i^*(dy)=cos(\theta) d\theta$
Best Answer
The vector field $\frac\partial{\partial\phi}$ is rotation invariant, hence the action generated by it must be rotation invariant. The only ortation invariant actions on $S^1$ are - rotations.