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Let X be a smooth manifold and $v, w \in \Gamma(TX)$ be two vector fields. Under the natural correspondence of vector fields and derivations on $C^\infty(X)$ let $\delta, \epsilon: C^\infty(X) \rightarrow C^\infty(X)$ be the corresponding derivations.

Then the commutator $[\delta, \epsilon] = \delta \circ \epsilon - \epsilon \circ \delta$ is another derivation. Define the Lie bracket $[v,w]$ as the vector field corresponding to this commutator. Using local coordinates on X, we can express the action of the two derivations on some $a \in C^\infty(X)$ as $$\delta(a) = \sum v_i \frac{\partial a}{\partial x_i} \ \text{ and } \ \epsilon(a) = \sum w_i \frac{\partial a}{\partial x_i}$$

Using this how can $[v,w]$ be expressed using the same local coordinates? I suspect the answer is the following, but I don't know how to derive it.

$$[v,w] = \sum_{i,j} \bigg(v_i \frac{\partial w_j}{\partial x_i}- w_i \frac{\partial v_j}{\partial x_i}\bigg) \frac{\partial}{\partial x_j}$$

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Your suspicion is correct. In the same local coordinates you have \begin{equation*} \begin{aligned} \ [v,w](a)&=\delta\circ \epsilon(a)-\epsilon\circ \delta(a)\\ &=\Big(\sum_i v_i\frac{\partial}{\partial x_i}\Big(\sum_j w_j\frac{\partial a}{\partial x_j}\Big)\Big)-\Big(\sum_j w_j\frac{\partial}{\partial x_j}\Big(\sum_i v_i\frac{\partial a}{\partial x_i}\Big)\Big)\\ &=\sum_{i,j} \Big(v_i\frac{\partial w_j}{\partial x_i}\frac{\partial a}{\partial x_j}+v_iw_j\frac{\partial^2a}{\partial x_ix_j}\Big)-\sum_{j,i} \Big(w_j\frac{\partial v_i}{\partial x_j}\frac{\partial a}{\partial x_i}+w_jv_i\frac{\partial^2a}{\partial x_jx_i}\Big)\\ &=\sum_{i,j}\Big(v_i\frac{\partial w_j}{\partial x_i}\frac{\partial a}{\partial x_j}-w_j\frac{\partial v_i}{\partial x_j}\frac{\partial a}{\partial x_i}\Big)+\sum_{i,j} \Big(v_iw_j\frac{\partial^2a}{\partial x_ix_j}-w_jv_i\frac{\partial^2a}{\partial x_jx_i}\Big)\\ &=\sum_{i,j}\Big(v_i\frac{\partial w_j}{\partial x_i}-w_i\frac{\partial v_j}{\partial x_i}\Big)\frac{\partial a}{\partial x_j}, \end{aligned} \end{equation*} where we smoothness of $a$ to cancel $\frac{\partial^2a}{\partial x_ix_j}-\frac{\partial^2a}{\partial x_jx_i}=0.$ Therefore $$[v,w]=\sum_{i,j}\Big(v_i\frac{\partial w_j}{\partial x_i}-w_i\frac{\partial v_j}{\partial x_i}\Big)\frac{\partial}{\partial x_j}.$$

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Associating the vector field $v_i$ with the differential operator $v:=v_i\partial^i$ gives $vwf=v_i\partial^i(w_j\partial^jf)$ for a sufficiently nice function $f$. The product rule, plus the symmetry of derivatives to delete the second-order terms, gives $[v,\,w]f=Rf$, where $R$ is the expression for $[v,\,w]$ you desire. The rest is an exercise.

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