Understanding the expression for Poisson bracket in local coordinates

Let $$M$$ be Poisson manifold with Poisson bracket $$\{\cdot,\cdot \}.$$ Let $$f,g \in C^{\infty} (M).$$ Let $$(U,x^i)$$ be a local coordinate of $$M.$$ Then it is claimed in my book that $$\{f,g\} \rvert_{U} = \sum\limits_{i,j} \pi_{ij} \frac {\partial f} {\partial x^i} \frac {\partial g} {\partial x^j}$$where $$\pi_{ij} = \{x^i, x^j\}.$$

Could anyone please explain to me as to where does the expression come from? Any help would be warmly appreciated.

For some point $$p = (p_1,\dots,p_n)$$, choose coordinates so that $$p = (0,0, \dots, 0)$$. Now approximate $$f$$ and $$g$$ by Taylor series: $$f(x) = f(0) + \left( \sum_i \frac{\partial f}{\partial x_i}(0) \cdot x_i \right) + F(x)$$ $$g(x) = g(0) + \left( \sum_j \frac{\partial g}{\partial x_j}(0) \cdot x_j \right) + G(x)$$ where $$F(x)$$ and $$G(x)$$ are the higher order terms (of degree $$\geq 2$$).
Now use these expressions to compute $$\{f,g\}$$. Since $$f(0)$$ and $$g(0)$$ are constants, they can be ignored (since constants are Casimirs), and since $$\frac{\partial f}{\partial x_i}(0)$$ and $$\frac{\partial g}{\partial x_j}(0)$$ are constants, they can be factored out: \begin {align*} \{f,g\}(x) &= \sum_{i,j} \frac{\partial f}{\partial x}_i(0) \frac{\partial g}{\partial x_j}(0) \{x_i, x_j\} \\ &\phantom{=} + \sum_i \frac{\partial f}{\partial x_i}(0) \{x_i, G\} + \sum_j \frac{\partial g}{\partial x_j}(0) \{F, x_j\} \\ &\phantom{=} + \{F,G\} \end {align*}
Notice that the first line of this equation is exactly what you want. So you need to check that the terms on the second and third line vanish when $$(x_1,x_2, \dots, x_n) = (0,0,\dots,0)$$. Remember that $$F(x)$$ and $$G(x)$$ are power series with all terms of degree $$\geq 2$$. So when you use the Leibniz rule, every term will have a factor of some $$x_k$$, so when $$x \to 0$$, all these terms will be zero.