# Closure under Lie Bracket — how is $c''(0)$ promoted to $(f\circ c)''(0)$

I've seen numerous different proofs that the tangent space to a Lie group is closed under $$[\cdot,\cdot]$$, i.e. that the Lie Bracket of two derivations is a derivation -- e.g. considering and differentiating the curve $$e^{\sqrt{t}X}e^{\sqrt{t}Y}e^{-\sqrt{t}X}e^{-\sqrt{t}Y}$$, or just showing that $$[D_1,D_2]$$ follows the product rule.

But one derivation I don't get comes from Timothy Goldberg's set of lecture notes The Lie Bracket and the Commutator of Flows. Here's the process:

1. Define the curve $$c(t)=\Phi_X^t\Phi_Y^t\Phi_X^{-t}\Phi_Y^{-t}(e)$$.
2. Show that $$[X,Y]=\frac12c''(0)$$.
3. Define an operation $$D:f(t)\mapsto (f\circ c)''(0)$$.
4. Show that $$D$$ is a derivation.

It's Step 3 I don't get. How do we know this operator $$D$$ is what "upgrades" $$[X,Y]$$ into a vector field? How can we show that $$[X,Y]$$ is the direction in which $$D$$ differentiates $$f$$?

Ah, never mind, it's obvious -- I just got confused because it's not true for all curves. Given $$(f\circ c)''(t)$$, it's clearly equal to
$$c''(t)\cdot\nabla f(t)+c'(t)\cdot\frac{d}{dt}\nabla f(t)$$
And since $$c'(0)=0$$ for the given curve, this is just equal to the first term.