$\newcommand{\g}{\mathfrak{g}} \newcommand{\Cinf}[1]{\mathcal{C}^\infty (#1)} \renewcommand{\d}{\mathrm{d}} \newcommand{\R}{\mathbb{R}} $
Let $\g$ be a Lie algebra and consider in $\g^*$ the following Poisson structure: $$\{f, g \} (\xi):= [f_{*,\xi},g_{*,\xi}](\xi), \qquad f,g \in \Cinf{\g^*},$$ where $f_{*,\xi},g_{*,\xi}: T_\xi \g^* = \g^* \to \R$ are considered as elements in $\g^{**}= \g$.
Given a Poisson map $\mu: M \to \g^*$ where $(M, \omega)$ is a symplectic manifold (therefore also a Poisson manifold) and $\g^*$ has the previous Poisson structure, consider also $\mu$ also as the map $\mu : \g \to \Cinf M, v \to \mu_v $ (linear in $v$), with $$\mu (x)(v)= \mu_v (x)$$
I want to show that $$\mu_{[u,v]}= \{\mu_u, \mu_v \}$$
Any idea, suggestion? This is part of a bigger problem but actually I am stuck here. Thanks.
Do you mean to say that $\mu: M\to \mathfrak{g}^\ast$ is a moment map? In fact, what you want to show is only true if the map $\mu$ is equivariant. If $\mu$ is a moment map, then $$d\mu([u,v])=\iota_{[u,v]}\omega=-L_u\iota_{v}\omega +\iota_{v}L_u\omega=-L_u\iota_{v}\omega.$$ You can show that this is equal to $d\{\mu(u),\mu(v)\}$.
It turns out that the constant $\mu([u,v])-\{\mu(u),\mu(v)\}$ measures exactly the equivariance of $\mu$. In particular, it is zero if $\mu$ is equivariant.