Lie bracket of vector fields - what have I done wrong? Let $X$ be a differentiable manifold and $U \subseteq X$ be an open subset with a chart $\phi: V \to U$ where $V \subseteq \mathbb{R}^n$. Let $v,w$ be vector fields on $X$ so that on $U$ we may write them $v = v_1 \frac{\partial}{\partial x_1}+...+v_n \frac{\partial}{\partial x_n}$, where $v_1,..,v_n: U \to \mathbb{R}$ are smooth and similarly for $w$. Here we understand $\frac{\partial}{\partial x_i}(b)(x)$ to mean the derivative of $b \circ \phi$ with respect to the i-th coordinate in $\mathbb{R}^n$ evaluated at $\phi^{-1}(u)$ where $b: U \to \mathbb{R}$ is smooth and $u \in U$. Then we define the Lie bracket by
$$[v,w] = \sum_{i=1,j=1}^{n} v_i \frac{\partial w_j}{\partial x_i}\frac{\partial}{\partial x_j} - w_j \frac{\partial v_i}{\partial x_j}\frac{\partial}{\partial x_i}$$
We want to check that this definition is independent of the choice of chart $\phi$. In order to better understand this, I tried to calculate a basic example.
Let $U=X=\mathbb{R}$ and $v_1,w_1,b: \mathbb{R} \to \mathbb{R}$ be given by $v_1(t)=1,b(t)=t,w_1(t)=t$. Now choose charts
$$x: \mathbb{R} \to U=\mathbb{R}; t \mapsto t  $$ and
$$y: \mathbb{R} \to U=\mathbb{R}; t \mapsto 2t  $$
Then $b(x(t))=t$ so $\frac{\partial b}{\partial x} = 1$, $v_1(x(t))=1$ so $\frac{\partial v_1}{\partial x} = 0$ and $w_1(x(t))=t$ so $\frac{\partial w_1}{\partial x} = 1$. Thus, for any $u \in U$, we have
$$[v,w](b)(u) = v_1(u) \frac{\partial w_1}{\partial x}(u)\frac{\partial b}{\partial x}(u) - w_1(u)\frac{\partial v_1}{\partial x}(u)\frac{\partial b}{\partial x}(u) = 1\times 1\times 1-u\times 0\times 1 = 1$$
On the other hand, $b(y(t))=2t$ so $\frac{\partial b}{\partial y} = 2$, $v_1(y(t))=1$ so $\frac{\partial v}{\partial y} = 0$ and $w_1(y(t))=2t$ so $\frac{\partial w_1}{\partial y} = 2$. Thus, for any $u \in U$, we have
$$[v,w](b)(u) = v_1(u) \frac{\partial w_1}{\partial y}(u)\frac{\partial b}{\partial y}(u) - w_1(u)\frac{\partial v_1}{\partial y}(u)\frac{\partial b}{\partial y}(u) = 1\times 2\times 2-u\times 0 \times 2 = 4$$.
It seems that here the Lie bracket does depend on the choice of chart. What have I done wrong?

Edited to make the distinction between $w_1$ and $w$ clear and to make clearer what definition of $\partial/\partial x_i$ I'm using.
 A: Informally, we have
$$
\frac{\partial~~}{\partial y} = \frac{\partial ~~}{\partial (2t)} = \frac{1}{2}\frac{\partial~~}{\partial t} = \frac{1}{2}\frac{\partial~~}{\partial x},
$$
and
\begin{align}
v &= 1 = \frac{\partial~~}{\partial x} = 2 \frac{\partial~~}{\partial y},\\
w &= t = t \frac{\partial~~}{\partial x} = 2t\frac{\partial}{\partial y}.
\end{align}
Therefore, in the first chart, we have
$$
[v,w] = \left(1 \frac{\partial t}{\partial x} - t \frac{\partial 1}{\partial x} \right) \frac{\partial~~}{\partial x} = (1 \times 1 - t \times 0)\frac{\partial~~}{\partial x} = \frac{\partial~~}{\partial x},
$$
since $\displaystyle \frac{\partial t}{\partial x} = 1$ (informally, "$x=t$").
In the second chart, we have
$$
[v,w] = \left(2\frac{\partial(2t)}{\partial y} - 2t\frac{\partial 2}{\partial y}\right) \frac{\partial~~}{\partial y} = (2 \times 1 - 2t\times 0)\frac{\partial ~~}{\partial y} = 2\frac{\partial ~~}{\partial y},
$$
since $\displaystyle \frac{\partial (2t)}{\partial y} = 1$ ( informally, "$y=2t$").
This is where you made a mistake: you seem to have considered that $\frac{\partial(2t)}{\partial y} = 2$.
Recall that we have $\displaystyle \frac{\partial~~}{\partial y} = \frac{1}{2}\frac{\partial~~}{\partial x}$, and thus there is no contradiction.

If you want to make the very first argument less informal, notice that if $\varphi\colon U \to \Bbb R^n$ is a chart, then the induced tangent frame is defined as
$$
\left.\frac{\partial~~}{\partial x^j}\right|_p = \left[d\varphi_p\right]^{-1} e_j,
$$
with $e_j$ the $j$th vector of the canonical basis of $\Bbb R^n$.
Therefore, if you canonically identify $T_t\Bbb R$ with $\Bbb R$, then
$$
\frac{\partial~~}{\partial x} = \mathrm{Id}^{-1} (1) = 1,
$$
while
$$
\frac{\partial~~}{\partial y} = \left[ 2\mathrm{Id}\right]^{-1}(1) = \frac{1}{2},
$$
hence the equality $\displaystyle \frac{\partial~~}{\partial x} = 2\frac{\partial~~}{\partial y}$.
Recall that $1 \in T_t\Bbb{R}$ corresponds to the derivation $D_1\colon f \mapsto f'(t)$, so that
$$
\frac{\partial (2t)}{\partial y} = \frac{1}{2}D_1(2t) = \frac{1}{2} (t\mapsto 2t)' = \frac{1}{2}\times 2 = 1. 
$$
Hence, for any $f$, the computations in the first chart gives
$$
\forall t\in \Bbb R,\quad \left([v,w]f\right)(t) = \left( 1 \times 1 - t \times 0\right) f'(t)=f'(t),
$$
and those in the second chart give
$$
\forall t\in \Bbb R,\quad \left([v,w]f\right)(t) = \left( 2 \times 1 - 2t \times 0\right) \frac{1}{2}f'(t) = f'(t).
$$
