This was essentially addressed in this question. Nick's comment suggests to use that $\left\{f,g \right\}=\sum_{i,j} \frac{\partial f}{\partial x^i} \frac{\partial g}{\partial x^j} \left\{ x_i, x_j \right\}$ on some coordinate neighborhood, but I can't figure out why this is.
I know that $\left\{ f, \cdot \right\}$ and $\left\{ \cdot, g \right\}$ are both derivations, so $X_f(g) = X_f(x_j) \frac{\partial g}{\partial x^j}$. So I'd like to show that $X_f(\cdot) = \left\{ \frac{\partial f}{\partial x^i} x_i, \cdot \right\}$. I get that, if $df=df_1$ on some neighborhood, then $f-f_1$ is a constant on that neighborhood and since $\left\{\cdot , g \right\}$ is a derivation for any $g$, we have that $\left\{f-f_1, g \right\} = 0$ for any $g$ on the coordinate neighborhood. I'd like to use this fact, but I don't see why we necessarily have that $df = d \left( \frac{\partial f}{\partial x^i} x_i \right)$, because the right hand side is $ df + d \left( \frac{\partial f}{\partial x^i} \right)x_i$.
As you remark $df$ and $d(\sum_i \partial_i f x_i)$ are not the same. Indeed you have in general $$\{f,g\}_x\neq \sum_i \{\partial_i f \,x_i,g\}_x$$ Remember also that you cannot pull out the $\partial_if$ without getting a $x_i\{\partial_i f,g\}_x$ term since this is a derivation.
What you do have is: $$\{f,g\}_x= df(\{\cdot ,g\}_x)=\sum_i \partial_i f\ dx_i(\{\cdot ,g\}_x) = \sum_i \partial_if\ \{x_i,g\}_x$$