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The question
Let $ M $ be a smooth manifold. Let $ \mathscr C_M^\infty $ be the sheaf of $ \mathscr C^\infty $-functions on $ M $, and let $ \mathscr X $ be the $ \mathscr C_M^\infty $-module of vector fields on $ M $, i.e. $$ \mathscr X(U) = \Gamma(U,\mathrm TM) = \{\text{sections $ X\colon U\to \mathrm TM $}\} $$ for any $ U\subset M $ open, where of course $ \mathrm TM $ is the tangent bundle of $ M $.
The dual $ \mathscr C_M^\infty $-module $ \mathscr X^* $ of $ \mathscr X $ is defined by $$ \mathscr X^*(U) = \underline\hom_{\mathscr C_M^\infty}(\mathscr X,\mathscr C_M^\infty)(U) = \{\text{morphisms of $ \mathscr C_M^\infty $-modules $ \mathscr X{\restriction_U}\to \mathscr C_U^\infty $}\} $$ for any $ U\subset M $ open, where $ \mathscr X{\restriction_U} $ is the restriction of the sheaf $ \mathscr X $ to the open subsets of $ U $.
I then took $ U = M $ and I tried to show that $$ \mathscr X^*(M) \cong \mathscr X(M)^* $$ where $ \mathscr X(M)^* $ is the dual $ \mathscr C_M^\infty(M) $-module of the $ \mathscr C_M^\infty(M) $-module $ \mathscr X(M) $, but I didn't succeeded.
An obvious map $$ \mathscr X^*(M) \rightarrow \mathscr X(M)^* $$ is the one that takes $ \omega = (\omega_U)_{\text{$ U\subset M $ open}}\in \mathscr X^*(M) $ to $ \omega_M\in \mathscr X(M)^* $.
I thought that to define a map in the opposite direction $$ \mathscr X^*(M) \leftarrow \mathscr X(M)^* $$ one could start with an $ \omega\colon \mathscr X(M)\to \mathscr C_M^\infty(M) $ and define $$ \omega_U(X) = \omega(\tilde X) $$ for all $ X\in \mathscr X(U) $, where $ \tilde X $ is something like an "extension" of $ X\colon U\to \mathrm TM $ to all of $ M $. I think that such a $ \tilde X $ could be defined using partitions of unity, but I'm not sure about that so I'm asking here for help.