It can be shown that the Hochschild cohomology of the algebra of smooth function on $X$ is naturally isomorphic to the sheaf of de Rham currents.
But is it possible to see from the very definition of Hochschild cohomology that this is a sheaf?
The Hochschild complex is given by continuous $(n + 1)$-linear forms on the space $C^\infty(X)$: $C^n = \{ \phi \mid C^\infty(X)^{n + 1} \to \mathbb{C}$} and with differential $b \colon C^{n-1} \to C^{n}$ given by $$ b\phi(f_0, \dotsc, f_n) = (-1)^n \phi(f_n f_0, f_1, \dotsc) + \sum_{0 \leq i < n} \phi(\dotsc, f_i f_{i+1}, \dotsc) \,. $$
Thanks