Translating between two different definitions of exterior derivative

Question

If $\Omega^k(M)$ is the space of differential $1$-forms on a manifold $M$, one may define the operator $\mathop{}\!\mathrm{d} : \Omega^k(M) \to \Omega^{k+1}(M)$ in a coordinate-independent way as follows: $$\begin{split} \mathop{}\!\mathrm{d}\omega(X_0,\dots,X_k) = &\sum_{i=0}^k (-1)^i X_i\left(\omega(\dots,\hat X_i,\dots)\right) \\ &+ \sum_{0\leq i < j \leq k} (-1)^{i+j}\omega([X_i,X_j],\dots,\hat X_i, \dots, \hat X_j,\dots) \end{split}\tag{1}$$ I am trying to see if I can recover the usual definition in terms of local coordinates $x$: $$\mathop{}\!\mathrm{d}\omega(X_0,\dots,X_k) = \frac{\partial \omega_{i_1\cdots i_k}}{\partial x^{i_0}}(\mathop{}\!\mathrm{d}x^{i_0} \wedge \mathop{}\!\mathrm{d}x^{i_1} \wedge \cdots \wedge \mathop{}\!\mathrm{d}x^{i_k}) (X_0,\cdots,X_k)\tag{2}$$ In the case $k = 1$, equations $(1)$ and $(2)$ become $$\begin{split} d\omega(X_0,X_1) &= X_0(\omega(X_1)) - X_1(\omega(X_0)) -\omega([X_0,X_1]) = \square \\ d\omega(X_0,X_1) &= \dfrac{\partial \omega_a}{\partial x^k} (\mathop{}\!\mathrm{d}x^k \wedge \mathop{}\!\mathrm{d}x^a)(X_0,X_1) = \triangle \end{split}$$ Setting $\omega = \omega_a \mathop{}\!\mathrm{d}x^a$, $X_0 = X_0^i \dfrac{\partial}{\partial x^i}$, $X_1= X_1^j \dfrac{\partial}{\partial x^j}$, we see that the first equation becomes $$\begin{split} \square &= X_0^i \dfrac{\partial}{\partial x^i}\left(\omega_a \mathop{}\!\mathrm{d}x^a \left(X_1^j \dfrac{\partial}{\partial x^j}\right)\right) - X_1^j \dfrac{\partial}{\partial x^j}\left(\omega_a \mathop{}\!\mathrm{d}x^a \left(X_0^i \dfrac{\partial}{\partial x^i}\right)\right) \\ &\quad - \omega_a \mathop{}\!\mathrm{d}x^a\left(\left(X_0^i \dfrac{\partial X_1^j}{\partial x^i} - X_1^j \dfrac{\partial X_0^i}{\partial x^j}\right)\dfrac{\partial}{\partial x^m}\right) \\ &= X_0^iX_1^j \dfrac{\partial \omega_j}{\partial x^i} - X_0^i X_1^j \dfrac{\partial \omega_i}{\partial x^j} - X_0^i \dfrac{\partial X_1^j}{\partial x^i} \omega_j + \dfrac{\partial X_0^i}{\partial x^j} X_1^j \omega_i \\ &= X_0^iX_1^j \left(\dfrac{\partial \omega_j}{\partial x^i} - \dfrac{\partial \omega_i}{\partial x^j}\right) - X_0^i \dfrac{\partial X_1^j}{\partial x^i} \omega_j + \dfrac{\partial X_0^i}{\partial x^j} X_1^j \omega_i \end{split}$$ On the other hand, the second equation becomes $$\begin{split} \triangle &= \dfrac{\partial \omega_a}{\partial x^k} X_0^i X_1^j \delta^{ka}_{ij} = \dfrac{\partial \omega_a}{\partial x^k} X_0^i X_1^j \det \begin{pmatrix} \delta^k_i & \delta^k_j \\ \delta^a_i & \delta^a_j\end{pmatrix}\\ &= \dfrac{\partial \omega_a}{\partial x^k} X_0^i X_1^j (\delta^k_i\delta^a_j - \delta^k_j\delta^a_i) = \left(\dfrac{\partial \omega_j}{\partial x^i} - \dfrac{\partial \omega_i}{\partial x^j}\right) X_0^i X_1^j \end{split} $$ This means that if $\square = \triangle$ then $$- X_0^i \dfrac{\partial X_1^j}{\partial x^i} \omega_j + \dfrac{\partial X_0^i}{\partial x^j} X_1^j \omega_i = 0 $$ Yet if this is so (and I still can't see why) then I'm at a loss at understanding the need for the second summation in equation $(1)$. Would someone care to elucidate and/or point out errors in my calculations?