I am following Hitchlin notes on differentiable manifolds:
Now suppose $Y$ is a vector field, considered as a map $Y : M → TM$. With a diffeomorphism $F : M → M$, its derivative $DF_x : T_x → T_{F(x)}$ gives $DF_x(Y_x) ∈ T_{F(x)}$
This defines a new vector field $\tilde{Y}$ by $$Y_{F(x)} = DF_x(\tilde{Y}_x) \ \ \ \ \qquad (6) $$ Thus for a function f, $$(\tilde{Y})(f ◦ F) = (Y f) ◦ F \qquad \quad (7)$$ Now if $F = ϕ_t$ for a one-parameter group (add: corresponding to a vector field X), we have $\tilde{Y}_t$ and we can differentiate to get $$Y˙ = \frac{∂} {∂t}\tilde{Y}_t \left. \right|_{t=0}$$ From (7) this gives $Y˙f + Y(Xf) = XY f$ so that $Y˙ = XY − Y X$ is the natural derivative defined above.
I don't understand the last sentence: how the equation 7 gives us this relation? If I use the flow, I have $$XYf = \frac{∂} {∂t} (Y(f) \circ \phi_t) \left. \right|_{t=0}$$ But I don't see how it is linked to the sum in the last sentence. Can someone helps me understand this?
Use the flow, I think you should have $$ \frac{∂} {∂t}\tilde{Y}_t \left. \right|_{t=0} (f) = \lim_{t\to0} \frac{Y(f\circ \phi_{-t}) \circ \phi_{t} - Y(f)}{t}. $$