Differential of a one-form eating a vector?

Since ${\rm d}(\alpha(X))(Y) = Y(\alpha(X))$, we indeed get from $${\rm d}\alpha(X,Y) = X(\alpha(Y)) - Y(\alpha(X)) - \alpha([X,Y])$$that $${\rm d}(\alpha(X))(Y) = X(\alpha(Y)) - \alpha([X,Y]) - {\rm d}\alpha(X,Y),$$but this is not very satisfying as we still have $X(\alpha(Y))$ on the right side. Since $\nabla$ is torsion free, we may improve the right side to obtain $${\rm d}(\alpha(X))(Y) = (\nabla_X\alpha)(Y) + \alpha(\nabla_YX) - {\rm d}\alpha(X,Y),$$where $(\nabla_X\alpha)(Y) = X(\alpha(Y)) - \alpha(\nabla_XY)$ is the covariant derivative of $\alpha$ in the direction of $X$. So if $\iota_X$ denotes interior product, we can write $$ {\rm d}(\alpha(X)) = \nabla_X\alpha + \alpha \circ \nabla X - \iota_X({\rm d}\alpha), $$which is as coordinate-free as we can get. I'm not sure how this would actually be useful in practice, though. One can do the same and get more complicated formulas for $k$-forms with $k>1$ by using the same strategy, by using that if $\omega$ is a $k$-form, then $\nabla_X\omega$ is a $k$-form with $$(\nabla_X\omega)(X_1,\ldots, X_k) = X(\omega(X_1,\ldots,X_k)) - \sum_{i=1}^k \omega(X_1,\ldots, \nabla_XX_i,\ldots, X_k).$$You will also have to use $$\begin{align} {\rm d}\omega(X_0,\ldots,X_k) &= \sum_{i=0}^k (-1)^k X_i(\omega(X_0,\ldots, \widehat{X_i},\ldots, X_k)) \\ &\qquad + \sum_{0\leq i<j\leq k} (-1)^{i+j} \omega([X_i,X_j],X_1,\ldots, \widehat{X_i},\ldots, \widehat{X_j},\ldots, X_k). \end{align}$$

You can use Cartan's magical formula $$ \mathcal{L}_X \omega = i_X d\omega + d(i_X \omega), $$ where $\omega$ is an $n$-form, $\mathcal{L}$ is the Lie derivative and $i$ stands for the interior product. Note that all these operations and the exterior derivative make sense on a differentiable manifold (not necessarily a Riemannian manifold) so one does not need the Levi-Civita connection in order to get a coordinate free expression. However, I believe there are formulas that relate the Lie derivative with the Levi-Civita connection, so you can use these if you want expressions using the Levi-Civita connection.

For the $1$-form $\alpha$ this gives: $$ d(\alpha(X)) = d(i_X \alpha) = \mathcal{L}_X \alpha - i_X d\alpha. $$

For the $2$-form $\beta$ one can repeatedly apply Cartan's formula: $$ \begin{align*} d(\beta(X,Y)) &= d(i_Y i_X \beta) \\ &= \mathcal{L}_Y(i_X \beta) - i_Y d(i_X \beta)\\ &= \mathcal{L}_Y(i_X \beta) - i_Y \mathcal{L}_X \beta + i_Y i_X d\beta. \end{align*} $$