For a Lie derivative $\mathscr{L}_{X} Y$ of $Y$ with respect to $X$, we mean that for two smooth vector fields $X$ and $Y$ on a smooth manifold $M$ such that the following holds $$ \mathscr{L}_{X} Y = \underset{t \rightarrow 0}{\lim} \frac{(\varphi_{-t})_{\star} Y - Y}{t}, \tag{1} $$ where $\varphi_{t}$ is a local one-parameter transformation group generated by $X$.
Suppose $(-\varepsilon, \varepsilon) \times U$ is the domain of $\varphi_{t}(p)$, where $U$ is an open subset of $M$. Then we can rewrite $(1)$ as the form $$ (\mathscr{L}_{X} Y)_{p} = \underset{t \rightarrow 0}{\lim} \frac{(\varphi_{-t})_{\star} Y_{\varphi_{t}(p)} - Y_p}{t}. \tag{2} $$
We regard the tangent vector $(\varphi_{-t})_{\star} Y_{\varphi_{t}(p)}$ as a map from an open subset of $(-\varepsilon, \varepsilon)$ into the tangent space $T_p M$.
My question is:
$1$. For an appropriate domain of $F$, why this map $F: t \mapsto (\varphi_{-t})_{\star} Y_{\varphi_{t}(p)}$ is smooth ? However, the mooth structure on $T_p M$ has been still unclear to me. Thanks in advance.
Since $Y$ is smooth in the variable $p$, and now $Y_{\varphi_t(p)}$ (or maybe this notation: $Y_{p(t)}$) is just the restriction of $p$ to the integral curve, and therefore also smooth. Think of it as a composition of the smooth map $t\mapsto p(t)$ with $p\mapsto Y_p$.
On the other hand, $(\varphi_{-t})_\star$ is nothing but a jacobian (linear map) parameterized by $t$, which is smooth since $\varphi_{-t}$ is.
It suffices to consider the smoothness of $t\mapsto (\varphi_{-t})_\star Y_{\varphi_t(p)}$ in local coordinates, which is the multiplication of a matrix corresponding to $(\varphi_{-t})_\star$ and a column vector corresponding to $Y_{\varphi_t(p)}$, which are both smooth in $t$.