It was left to me as an exercise to show the naturality of the Lie derivative of arbitrary tensors on a smooth manifold. Is the following argument correct (it seems too easy)?
Let $X$ be a vector field and let $T$ be a tensor on a smooth manifold $N$. Let $\theta^p(t)$ be the flow associated to $X$ such that $\theta^p(0)=p$ and $\theta'^p(0)=X.$
Consider an arbitrary diffeomorphism $\phi$ from a smooth manifold $M$ to $N$, such that $\phi(x)=p.$ I need to show $\phi^*(L_XT)= L_{\phi^*X} \phi^*T.$
We compute $\phi^* (L_X T)$ at the point $x$ \begin{align} \phi^*(L_X T)&= \phi^* \left(\lim_{t \to 0} \frac{1}{t} (\theta^p(t) T- T ) \right)\\ &= \lim_{t \to 0} \frac{1}{t} \left(\phi^* (\theta^p(t) T- T)\right) \\ &= \lim_{t \to 0} \frac{1}{t} (\phi^*(\theta^p(t)) T- \phi^*T))\\ &= L_{\phi^*X}\phi^*T. \end{align}
The above computation uses only the fact that $\phi^*$ commutes with limits, which is clear from writing everything out in coordinates and using the fact that the maps are all smooth.