Differential of the Rabinowitz Action Functional

Question

On an exact Hamiltonian system $(M,d\alpha,H)$ define the Rabinowitz action functional $$\mathcal{A}^H \colon C^\infty(\mathbb{S}^1,M) \times (0,+\infty) \to \mathbb{R}$$ by $$\mathcal{A}^H(\gamma,\tau) := \int_{\mathbb{S}^1}\gamma^*\alpha - \int_{\mathbb{S}^1}H(\gamma(t))dt.$$ Following the book The Restricted Three-Body Problem and Holomorphic Curves by Frauenfelder/Koert, p. 97 it is claimed that $$d\mathcal{A}^H_{(\gamma,\tau)}(X,\widehat{\tau}) = \int_{\mathbb{S}^1}\gamma^*\mathcal{L}_X\alpha - \tau \int_{\mathbb{S}^1}dH(X) - \widehat{\tau}\int_{\mathbb{S}^1}H \circ \gamma$$ for any pair $$(X,\widehat{\tau}) \in T_\gamma C^\infty(\mathbb{S}^1,M) \times \mathbb{R} = \Gamma(\gamma^*TM) \times \mathbb{R}.$$ I do not understand this computation. I mean, it is clear to me that to compute the differential (or merely define it in that infinite-dimensional setting), we pick a variation of $(\gamma,\tau)$, for example $$\Gamma(\varepsilon,t) := (\exp_{\gamma(t)}^\nabla(\varepsilon X_t), \tau + \varepsilon \widehat{\tau}).$$ Then $$d\mathcal{A}^H_{(\gamma,\tau)}(X,\widehat{\tau}) = \frac{d}{d\varepsilon}\bigg\vert_{\varepsilon = 0}\mathcal{A}^H(\Gamma(\varepsilon,\cdot)).$$ So in the first term we end up with $$\int_{\mathbb{S}^1}\frac{d}{d\varepsilon}\bigg\vert_{\varepsilon = 0}\gamma_\varepsilon^*\alpha,$$ where $\gamma_\varepsilon := \exp_\gamma^\nabla(\varepsilon X)$. What I do not understand is, why that $$\int_{\mathbb{S}^1}\frac{d}{d\varepsilon}\bigg\vert_{\varepsilon = 0}\gamma_\varepsilon^*\alpha = \int_{\mathbb{S}^1}\gamma^*\mathcal{L}_X\alpha.$$ I mean, since $X$ is only a vector field along $\gamma$, for me, the expression $\mathcal{ L}_X\alpha$ does not even make sense.