Let $M$ be a smooth manifold, $V$ a smooth vector field whose flows is denoted by $\theta$. For any smooth cotensor field $A$, we define the Lie derivative with respect to $V$ as $L_V A = (\frac{d}{dt} \theta_t^* A)(0)$. It is relatively easy to show that this definition implies the four properties bellow:
a) $L_V$ is linear over $\mathbb{R}$,
b) $L_V f = Vf$ for any $0$-tensor $f$,
c) $L_V(A \otimes B) = (L_V A) \otimes B + A \otimes (L_V B)$ for two smooth cotensor fields,
d) $L_V(w(X)) = (L_V w)(X) + w([V,X])$ for $w$ a rank 1 smooth cotensor field and $X$ a smooth vector field.
I would like to show the reciprocal. That is, an operator $L_V$ as above (acting on smooth cotensor fields) is indeed the Lie derivative in the sense of pullback by flow.
To do so, I first proved that $L_V$ is acting locally: for any smooth cotensor fields $A$ and $B$, if $A$ is equal to $B$ on a neighborhood of a point $p$ of $M$, then $(L_V A)_p = (L_VB)_p$. Because of this local property, it is enough to choose some coordinate system $(x^i)_{i = 1, \ldots, n}$ centered at some arbitrary $p$ and show that $(L_V A)_p$ coincides with the pullback definition above. One can first try to show it for the basis cotensors $dx^{i_1} \otimes \ldots \otimes dx^{i_k}$, but even in the case $k = 2$ I'm stuck. One probably has to use d) with $w = dx$, but maybe I'm on a totally wrong road.
Thanks for help.
Let's call the new one $\tilde L$ for the moment. You want to show $$\tag{1} \tilde L_V A = L_VA$$ for all vector fields $V$ and for all cotensor field $A$.
First by $(a)$, $(1)$ is true for all functions $A=f$. Using this fact and $(d)$, $(1)$ holds when $A$ is a differential one form $\omega$. Now we use your observation that $\tilde L, L$ depends only locally on $A$. So if $A$ is a $2$-tensor, then $A$ is locally written as $$A = \sum_{i,j=1}^n f_{ij} (x) dx^i \otimes dx^j.$$ Then using the observation for functions and one form, together with $(c)$, one can show $(1)$ for all $2$-tensor $A$. Then the general cases can be done inductively.