Let $M$ be a closed manifold and $\omega$ a closed $2$-form on $M$ such that $$ \int_{S^2} f^* \omega \in \mathbb Z $$ for any smooth $f \colon S^2 \to M$. For a given $x \in \mathcal LM = C^\infty(\mathbb R / \mathbb Z,M)$, assuming contractibility, we set $$ L(x) = \int_{B^2} u^*\omega \quad \in \mathbb R / \mathbb Z, $$ where $B^2 = \{ z \in \mathbb C \mid |z| \leq 1\}$ and $u \colon B^2 \to M$ is a smooth map such that $u(e^{2\pi i t}) = x(t)$. I want to show that $$ dL(x)\xi = \int_0^1 \omega(\xi(t),\dot x(t)), $$ for any $\xi \in C^\infty(x^* TM) = T_x \mathcal LM$.
I tried to do this by definition: I take a family of closed loops $x_s(t)$ such that $x_0(t) = x(t)$ and also $\partial_s x_s(t)|_{s=0} = \xi(t)$. Then I continue $\xi$ to some vector field aroung $u(B^2)$ and consider its flow $\phi_s$. Then I compute $$ \partial_s L(x_s(t))|_{s=0} = \int_{B^2} \partial_s (\phi_s u)^*\omega |_{s=0} = \int_{B^2} \partial_s u^* \phi_s^*\omega |_{s=0} \\ \int_{B^2} du^* \circ \partial_s \phi^*_s \omega |_{s=0} = \int_{B^2}du^* \circ \mathcal L_\xi \omega , $$ taking into account that $u_s = \phi_s u \colon B^2 \to M$, $u_s(e^{2\pi i t}) = x_s(t)$, and the definition of the Lie derivative. However, I don't see how to procced. Could you help me please?