Let $(M,g)$ be a Riemannian manifold and $T$ be a smooth tensor field. The official definition of the Lie derivative of $T$ with respect to a field $X\in\Gamma(TM)$ is: $$\mathcal{L}_X(T)_p:=\left.\frac{d}{dt}\right|_{t=0}(\varphi_{-t})_*T_{\varphi_t(p)}$$
(where $\varphi$ is the flow of $X$)
What does this notation $(\varphi_{-t})_*T_{\varphi_t(p)}$ mean?
Taking for example $T=g$, I believe it would make sense to define $\mathcal{L}_X$ as follows: $$\mathcal{L}_X(g)_p(Y_p,Z_p):=\left.\frac{d}{dt}\right|_{t=0}g_{\varphi_{t}(p)}((d\varphi_t)_p(Y_p),(d\varphi_t)_p(Z_p))$$
But I don't know how to reconciliate this with the $-t$ in the official definition.
First of all, Lie derivatives are defined on any smooth manifold, so the Riemannian structure is superfluous.
Let $T$ be a tensor field and $X$ a vector field. If you wish to look at the Lie derivative at a point $p$, $(\mathcal{L}_XT)_p$, you must look at the tensor at the point $\phi(t,p)$, where $\phi(t,p)$ is the flow of $X$ generated from $p$. You now need a way to push this tensor $T_{\phi_t(p)}$ from the point $\phi_t(p)$ back to the point $p$. Since $\phi(0,p)=p$, we have that $$\phi(-t,\phi(t,p))=\phi(-t+t,p)=\phi(0,p)=p.$$ So we use the pushforward of the function $\phi_{-t}$ at the point $\phi_t(p)$, that is, the map $d(\phi_{-t})_{\phi_t(p)}$ (I'm using this notation instead of the $*$ notation for subscript clarity). We then see that $$d(\phi_{-t})_{\phi_t(p)}T_{\phi_t(p)}$$ is now a tensor at $p$ for any defined $t$. We can now take the usual time derivative as normal since we're in a vector space, resulting in $(\mathcal{L}_XT)_p$.