Is the Lie bracket always invariant under coordinate transformations? This is my first question on StackExchange. I think it's probably quite easy, but it's been baffling me for a while.
I'm  doing computations to determine invariant properties of the quantity $X\circ Y= \nabla_Y X$ where $X$ and $Y$ are vector fields,  when we change coordinate choices (when $\nabla$ is a flat torsion-free connection).
So naturally, we start with the simplest case, $\mathbb{R}$ with its usual connection. However we encounter the following problem.
Start with vector fields $X=f(x)dx$ and $Y=g(x)dx.$ We have $X\circ Y = g(x)\frac{df}{dx}dx.$  We then make a general change of coordinates $\phi(y)=x.$ So $X=f(\phi(y))\frac{d\phi}{dy}dy$ and $Y=g(\phi (y))\frac{d\phi}{dy}dy.$ Then $X \circ' Y = g(\phi(y))\frac{d\phi}{dy}(\frac{df(\phi(y))}{d \phi}\frac{d \phi}{dy}+f(\phi(y))\frac{d^2\phi}{d y^2})dy$ in the new coordinates. Attempting to return back to $x$-coordinates we get $X \circ' Y =(\frac{d \phi}{dy})^2(g(x)\frac{df}{dx}dx)+ g(x)f(x)\frac{d^2 \phi}{dy^2}dx$. The rightmost term is symmetric in $X$ and $Y$ and for the purposes of this question can be neglected.
In this case $X\circ Y-Y\circ X = [Y,X]$ as $\nabla$ is flat torsion free. But in the expression above we pick up a factor of $(\frac{d\phi}{d y})^2$ when we compute the bracket. But the bracket is coordinate independent! What's gone wrong?

The mistake is in the change of coordinates. If $x=\phi(y)$, then
$$dx = d\phi = \frac{d\phi}{dy}(y)\,dy$$
is the change of coordinates for one-forms, not vectors. For vectors you have the dual formula
$$\frac{\partial}{\partial_x} = \left(\frac{d\phi}{dy}(y)\right)^{-1}\frac{\partial}{\partial_y}.$$
