I just derived that in local coordinates (it suffices to centre) around $0$, that
$$\{f,g\}(x)=\sum_{i,j}\{x^i,x^j\}\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}$$
only using the axiomatic properties of the Poisson bracket $$\{*,*\}:C^\infty(M)\times C^\infty(M)\to C^\infty(M)$$ 1) The conjugate of Poisson is the Poisson of conjugates, 2) bracket is bilinear, 3)antisymmetric, 4) Leibniz rule is satisfied, and 5) the Jacobi identity is satisfied.
Essentially I took the Taylor series around $0$ then the first order terms clearly die out ($\{f,1h\}=\{f,1\}h+\{f,h\}$, so any constant is destroyed by the bracket), and the higher order terms die out because you get
$$\sum_{ij} \frac{\partial^2 f}{\partial x^i \partial x^j}\frac{\partial^2 g}{\partial x^i \partial x^j}\{x_i x_j,x_i x_j\}$$ for the Poisson bracket of the second order terms, third order terms have $\{x^i x^j x^k,x^i x^j x^k\}$ and so on, and obviously $\{g,g\}=-\{g,g\}=0$.
My question is what is (geometrically or otherwise) the meaning of $\{x^i,x^j\}$? Again, $\{x^i,x^i\}=0$ but what should that $mean$.
The equation you got for the Poisson bracket shows that the functions $\{x^i,x^j\}$ piece together to give the coefficients for a global anti-symmetric tensor field (called a bivector field) $$ \Pi = \{x^i,x^j\} \frac{\partial}{\partial x^i} \wedge \frac{\partial}{\partial x^j} \in \Gamma(M, \Lambda^2 TM). $$ We recover the Poisson bracket by the equation $$ \{f,g\} = \Pi(df,dg). $$
Given any bivector field the above equation makes sense but may not be a Poisson bracket since the Jacobi identity could fail.
If the field $\Pi$ is non-degenerate (i.e. non-degenerate as a bilinear form $T_x^* M \times T_x^*M\to \mathbb R$) then we have an isomorphism $T^* M \to TM$ and we can convert $\Pi$ to a differential 2-form $\omega$. Then $\omega$ is a symplectic structure (closedness of $\omega$ is equivalent to the Jacobi identity for $\{\cdot,\cdot\}$).