I am currently working on the first variation formula of energy and I use this file here to understand this notion.. We are working on the following construction:
Let $\gamma : [a,b] \to M$ be a smooth curve, and $\epsilon >0$. A variation of $\gamma$ is a smoot map $f :[a,b] \times (-\epsilon, \epsilon)\to M$ so that $$f(t,0)=\gamma(t)$$ for all $t \in [a,b]$. In what follows, we will also denote $\gamma_s(t)=f(t,s)$.
Let $f$ be a varation of $\gamma$. For simplicity we will denote $$f_s:= df(\frac{\partial}{\partial s}), \;\;\;\;f_t:=df(\frac{\partial}{\partial t}) $$
By definition we have $f_t= \gamma^{'}_s$.
Why do we have this: $f_t= \gamma^{'}_s$? I mean we have that $\gamma'_s(t)= \frac{\partial}{\partial t} \gamma_s(t)= \frac{\partial}{\partial t} f(t,s)$. Why is this now equal to $df(\frac{\partial}{\partial t})$? What do I oversee? Many thanks for some help!
By definition, at each $(t_0, s_0) \in [a, b] \times (-\varepsilon, \varepsilon)$, $\left.\frac{\partial}{\partial t}\right\vert_{(t_0, s_0)} \cong (1, 0)$ and $\left.\frac{\partial}{\partial s}\right\vert_{(t_0, s_0)} \cong (0, 1)$ are tangent vectors living in $T_{(t_0, s_0)}([a,b] \times (-\varepsilon, \varepsilon)) \cong T_{(t_0, s_0)} \mathbb{R}^2$ which act on real smooth functions (actually germs of functions, but that's not important now) $h: U_{(t_0, s_0)} \to \mathbb{R} $ defined on a neighbourhood of $(t_0, s_0)$ by
$$\left.\frac{\partial}{\partial t}\right\vert_{(t_0, s_0)} (h) \doteq \left.\frac{\partial h}{\partial t}\right\vert_{(t_0, s_0)} = \lim_{\delta \to 0} \frac{h(t_0+ \delta, s_0) - h(t_0, s_0)}{\delta} = \mathrm{d}h_{(t_0, s_0)}((1, 0))$$
and $$\left.\frac{\partial}{\partial s}\right\vert_{(t_0, s_0)} (h) \doteq \left.\frac{\partial h}{\partial s}\right\vert_{(t_0, s_0)} = \lim_{\delta \to 0} \frac{h(t_0, s_0 + \delta) - h(t_0, s_0)}{\delta} = \mathrm{d}h_{(t_0, s_0)}((0, 1))$$
Similarly, at each $(t_0, s_0) \in [a, b] \times (-\varepsilon, \varepsilon)$, $\left. \frac{\partial f}{\partial t}\right\vert_{(t_0, s_0)}$ is defined by: $$\left. \frac{\partial f}{\partial t}\right\vert_{(t_0, s_0)} = \mathrm{d}f_{(t_0, s_0)}\left(\left.\frac{\partial}{\partial t}\right\vert_{(t_0, s_0)} \right) = \mathrm{d}f_{(t_0, s_0)}((1, 0))$$
Let us fix $s_0 \in (-\varepsilon, \varepsilon)$ and consider the curve $\gamma_{s_0}: [a, b] \to M$ given by $\gamma_{s_0}(t) = f(t, s_0)$. If we define the curve $\alpha_{s_0} : [a, b] \to [a, b] \times (-\varepsilon, \varepsilon)$ by $\alpha_{s_0}(t) = (t, s_0)$, then we have by the chain rule:
\begin{aligned} \gamma_{s_0}'(t) = (f \circ \alpha_{s_0})'(t) &= \mathrm{d}f_{\alpha_{s_0}(t)}(\alpha_{s_0}'(t)) \\ &= \mathrm{d}f_{(t, s_0)}((1, 0)) \\ &= \left. \frac{\partial f}{\partial t}\right\vert_{(t, s_0)} \end{aligned} as desired.