Let $M$ be a manifold, $\omega \in \Omega^k(M)$. For all vector fields $X_1,...,X_{k+1}$ on $M$, we define
\begin{align*} \Omega(X_1,..., X_{k+1}) &= \sum\limits_{i=1}^{k+1} (-1)^{i+1}X_i\big(\omega(X_1,..., \hat{X_i},..., X_{k+1})\big) \\ &\quad + \sum\limits_{i< j} (-1)^{i+j} \omega([X_i,X_j], X_1,....., \hat{X_i},..., \hat{X_j},.., X_{k+1}). \end{align*}
I already proved that $\Omega$ is multilinear over $C^\infty(M)$. Now let $p \in M$ and $v_1,...,v_{k+1} \in T_pM$ and for every $i \in \{1,...,k+1\}$ let $X_i$ be any vector field on $M$ with $(X_i)_p = v_i$. I now have to show that $\Omega(X_1,...,X_{k+1})$ only depends on $v_1,...,v_{k+1}$ and not on the choice of vector fields $X_i$ extending the vectors $v_i$ but I don't really have an idea on how to start.
This is a standard argument using $C^\infty$ multilinearity. Suppose you take $X_1$ and $X_1'$ both having value $v_1$ at $p$, and consider their difference $Z$. We want to verify that $\Omega(Z,X_2,\dots,X_{k+1})(p) = 0$.
Work in local coordinates $(x^1,\dots,x^n)$ centered at $p$ and write $Z=\sum a^j\frac{\partial}{\partial x^j}$. We know that $a^j(p)=0$ for all $i$. Now, let $\psi$ be a bump function with $\psi(p) = 1$. Here's the main trick: \begin{align*} \Omega(Z,\dots)(p) &= \Omega(\psi^2Z,\dots)(p) = \Omega(\sum(\psi a^j)(\psi\tfrac{\partial}{\partial x^j}),\dots)(p) \\ &= \sum (\psi a^j)(p)\Omega(\psi\tfrac{\partial}{\partial x^j},\dots)(p) = 0. \end{align*}