Showing that the global formula for the exterior derivative is $C^{\infty}$-linear with respect to $X_i$.

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For context, I'm trying to make sense of Exercise 1.123 from Gallot, Hulin, Lafontaine's Riemannian Geometry book.

I'm assuming (and have proved this is the past usign Tu's book) that for $\alpha\in \Omega^{p}(M)$ and $(X_0, \ldots, X_p)$ are $p+1$ vector fields on $M$, then $$d\alpha(X_0, \ldots, X_p)=\sum_{i=0}^{p}(-1)^{i}L_{X_i}(\alpha(X_0, \ldots, \hat{X_i}, \ldots, X_p))+\sum_{0\leq i<j\leq p}(-1)^{i+j}\alpha([X_i, X_j], X_o, \ldots \hat{X_i}, \ldots, \hat{X_j}, \ldots, X_p),$$ where the hat denotes an omitted term.

I want to show that the right hand side is $C^{\infty}(M)$-linear with respect to $X_i$.

This is my setup so far and I'm not sure that it's right either.

$$d\alpha(X_0, \ldots,fX_i, \ldots, X_p)=\sum_{i=0}^{p}(-1)^{i}L_{fX_i}(\alpha(X_0, \ldots, \hat{fX_i}, \ldots, X_p))+\sum_{0\leq i<j\leq p}(-1)^{i+j}\alpha([fX_i, X_j], X_o, \ldots \hat{fX_i}, \ldots, \hat{X_j}, \ldots, X_p),$$ $$=\sum_{i=0}^{p}(-1)^{i}fL_{X_i}(\alpha(X_0, \ldots, \hat{fX_i}, \ldots, X_p))+ (df\wedge i_{X_i}\alpha)(\alpha(X_0, \ldots, \hat{fX_i}, \ldots, X_p))+ \sum_{0\leq i<j\leq p}(-1)^{i+j}\alpha(fX_i X_j-X_j fX_i, X_o, \ldots \hat{fX_i}, \ldots, \hat{X_j}, \ldots, X_p).$$

How can I finish this or is my setup even correct?