Confused about intuition behind Lie derivative

Solution 1:

For the intuition behind the Lie derivative I can maybe add a bit to the comment by Qiaochu Yuan: The simplest way to understand the differentiation is to consider the flows $\phi_t$, use them to pull back $T$ and then evaluate in the point $p$. This lies in the fiber of the appropriate tensor bundle at $p$ for any value of $t$. Hence you obtain a smooth curve in a finite dimensional vector space, and you can simply differentiate this in $t=0$ to obtain an element of that vector space. Work would be needed to deduce that this depends smoothly on $p$, but for the purpose of intuition that's not neccesary.

This description also implies the answer to question 1, but I can give you a more direct argument: It is true that the Lie derivative is a local operator, so the value of $\mathcal L_XT$ in a point $p$ depends only on the restrictions of $X$ and $T$ to an arbitrarily small neighborhood $U$ of $X$. But much more is true, because the Lie derivative is a first order differential operator. This means that it depends only on the one-jets of $X$ and $T$ in $p$. (There are various ways to make this explicit.) If you look at the coordinate formula for the Lie derivative, you see that to compute $\mathcal L_XT(p)$ you only need the components of $T$ and their derivatives in direction $X(p)$. Hence it is indeed true that the value of $\mathcal L_XT$ in $p$ depends only on the restriction of $T$ to the flow line of $X$ through $p$.

Question 2 is anserwered in Jack Lee's comment.