Link between maps $C^\infty(M) \times C^\infty(M) \rightarrow C^\infty(M)$ and bivectors.

Question

I was reading a bit these notes on Poisson geometry.

At some point, the author establishes a connection between bi-vectors and maps $C^\infty(M) \times C^\infty(M) \rightarrow C^\infty(M)$ and says.

we can define a bivector field $\pi ∈ \mathfrak{X}^2(M)$ by:

$\pi(df,dg):=\{f,g\}$.

But how is $\pi$ defined for the 1-forms that are not exact?