Let $V$ be a $\mathbb{C}-$ vector space, $J$ an almost complex structure on $V$ and take a real orthonormal basis $\langle v_1,Jv_1,\ldots,v_n,Jv_n\rangle $ with a scalar product $\langle,\rangle = \sum_{i\leq m}v^*_i\otimes v^*_i + (Jv_i)^*\otimes (Jv_i)^*$, where $v^*_i$ are the dual.
We can define the fundamental form $\omega (v,w):=\langle Jv, w \rangle$.
It turns out that $\omega = \sum_{i\leq m}v^*_i\wedge (Jv_i)^* $.
I am trying to deduct that, but I fail to do so. I tried to calculate $\omega (v_i,v_j)$ but I am stuck here $\omega (v_i,v_j)=\sum_{i\leq m}(v^*_i\otimes v^*_i)(Jv_i\otimes v_j) + ((Jv_i)^*\otimes (Jv_i)^*)(Jv_i\otimes v_j)$
Where $(v^*_i\otimes v^*_i)(Jv_i\otimes v_j)=v^*_i(Jv_i)\cdot v^*_i(v_j)$ which gives $0$ when $i\neq j$ and $J$ $(v^*_i(Jv_i)= Jv^*_i(v_i ) $? when $i= j$. And similar for $(Jv_i)^*\otimes (Jv_i)^*)(Jv_i\otimes v_j)$ which doesn't make any sense.
Can someone explain what we mean and how to get $\omega = \sum_{i\leq m}v^*_i\wedge (Jv_i)^* $ ?
Your approach is mostly correct. In fact, we can compute $\omega(v, w)$ for all $v,w$ in a set of basis vectors to see that $\omega$ indeed has the proposed expression. But $(v_1, \dots, v_n)$ is not a complete real basis of $V$, so computing $\omega(v_i, v_j)$ is not enough. In fact, you also have to compute $\omega(v_i, J v_j)$, $\omega(Jv_i, v_j)$ and $\omega(Jv_i, Jv_j)$. To do this, note that $(Jv_i)^*$ is the unique element of the dual, which sends $Jv_i$ to $1$ and all other basis vectors to $0$. If you do this, you find:
$$\omega(v_j, v_k) = \sum_{i} \underbrace{{v_i}^*(Jv_j)}_{=0} \cdot {v_i}^*(v_k) + (Jv_i)^*(Jv_j) \cdot \underbrace{(Jv_i)^*(v_k)}_{=0} = 0$$
$$\omega(v_j, Jv_k) = \sum_{i} \underbrace{{v_i}^*(Jv_j)}_{=0} \cdot {v_i}^*(Jv_k) + \underbrace{(Jv_i)^*(Jv_j)}_{=\delta_{ij}} \cdot \underbrace{(Jv_i)^*(Jv_k)}_{=\delta_{ik}} = \delta_{jk}$$
and similarly $\omega(Jv_j, v_k) = -\delta_{jk}$ and $\omega(Jv_j,Jv_k) = 0$. For the latter two terms you have to use $J^2 = -\operatorname{id}_V$. This then immediately gives
$$\omega = \sum_{i} {v_i}^*\wedge (Jv_i)^*$$ as these two bilinear forms agree on a real basis of $V$.