I think I have some issues with the notation used in the proposition below
Let $X, Y$ be differentiable vector fields on a manifold $M$, let $p \in M$, and let $\varphi_t$ be the local flow of $X$ in a neighborhood $U$ of $p$ Then $$ [X,Y](p) = \lim_{t\to 0} [Y - d\varphi_t Y](\varphi_t(p)) $$
More specifically there's a limit at some point of the proof:
$$ \lim_{t \to 0} \frac{(Yf)(\varphi_t(p)) - (Yf)(p)}{t} = (X(Yf))(p) $$
which is the bit I don't quite get...
Could you help?
Update:
I was just reading again through this part, the following statement might actually help to understand the limit in my question
Let $X$ be a differentiable vector field on a differentiable manifold $M$, and let $p \in M$. Then there exist a neighborhood $U \subset M$ of $p$, an interval $(-\delta,\delta)$, $\delta > 0$ and a differentiable mapping $\varphi : (-\delta,\delta) \times U \to M$ such that the curve $t \to \varphi(t,q)$, $t \in (-\delta,\delta)$, $q \in U$ is the unique curve which satisfies $$ \frac{\partial \varphi}{\partial t} = X(\varphi(t,q)) $$ and $\varphi(0,q) = q$. A curve $\alpha:(-\delta,\delta) \to M$ which satisfies the condition $\alpha'(t) = X(\alpha(t))$ is called a trajectory of the field $X$ that passes through $q$ for $t = 0$
It is not explained exactly what does it mean that $\varphi(t,q)$ is differentiable, but given the definition I'd say that if we fix $q \in M$ then the restriction $\alpha_q(t)$ is curve on the manifold $M$, therefore I'd say that differentiable in this context would mean that the $\alpha'_q(0) \in T_q M$ exists, and this means
$$ \alpha'_q(0)f = \left. \frac{d}{dt}(f \circ \alpha_q) \right|_{t=0} = \left. \frac{\partial}{\partial t} (f \circ \varphi(\cdot,q)) \right|_{t=0} $$
exists, here $f$ is a differentiable function defined on the manifold $M$. Now if $Y$ is a differentiable vector field, which can be seen as an operator, instead of using $f$ I can use $Yf$, therefore I end up with
$$ \left. \frac{\partial}{\partial t} \left((Yf) \circ \varphi(\cdot,q) \right) \right|_{t=0} = \lim_{t \to 0} \frac{(Yf)(\varphi_t(p)) - (Yf)(p)}{t} $$
But it must be
$$ \frac{\partial \varphi}{\partial t} (Yf) = \left. \frac{\partial}{\partial t} \left((Yf) \circ \varphi(\cdot,q) \right) \right|_{t=0} $$
By definition of tangent vector, since $\varphi$ is a local flow for $X$ (in the proposition) we can write
$$ \frac{\partial \varphi}{\partial t} (Yf) = X(Yf)(\varphi_t(q)) $$
Does this make sense?
The equation you have displayed is wrong. What is correct is that $$\lim_{t\to 0}\frac{(Yf)(\phi_t(p)) - (Yf)(p)}t = X_p(Yf) = (X(Yf))(p).$$ (This follows because the curve $\phi_t(p)$ is a curve through $p$ with tangent vector $X_p$ at $p$.)