Lie derivative of a one form in $S^1$

First of all, it is important to notice that $\theta$ and $\varphi$ stand for the same thing here. In her paper, the authors refers to $\theta$ when $\Bbb S^1$ is thought of as the coordinate on the group $\Bbb S^1$, acting on a manifold $M$. She refers to $\varphi$ when it is thought of as the manifold $\Bbb S^1$. For the action by translation of the circle on itself, they coincide and hence, $J = -\frac{\partial}{\partial \theta} = - \frac{\partial}{\partial \varphi}$.

is every one form is of this form?

$\Omega^1(\Bbb S^1)$ is a $\mathcal{C}^{\infty}(\Bbb S^1)$-module of rank $1$. Since $d\varphi$ is nonzero everywhere, $\{d\varphi\}$ is a basis of $\Omega^1(\Bbb S^1)$, that is, any differential form of degree $1$ on the circle is of the form $gd\varphi$.

why does the equation $\mathcal{L}(J)[gd\varphi ]=0$ implies that $g$ is constant

Since $\mathcal{L}(J)$ is a derivation, $\mathcal{L}(J)[g d\varphi] = (\mathcal{L}(J)g)d\varphi + g \mathcal{L}(J) d\varphi$. You can check that $\mathcal{L}(J) d\varphi = 0$ since the action of $\mathbb{S}^1$ onto itself is by translation and hence the coordinate $\varphi$ is invariant under this action. Therefore, $\mathcal{L}(J)[gd\varphi] = 0 \iff \mathcal{L}(J)g = 0$. The problem reduces to solving the equation $\mathcal{L}(J)g = 0$, which is in fact just the equation $dg = 0$, and since $\Bbb S^1$ is connected, its solutions are constant maps.

Comment The author isn't a 'he', but a 'she', since Michèle Vergne is a woman although she's a mathematician ;)