The coordinate expression of the Lie derivative of a covector field, $w$ with respect to a vector field $X$ is given by:
$$ \mathcal{L}_{X}w = \left(X^{\alpha}\frac{\partial w_{\mu}}{\partial x^{\alpha}} + w_{\alpha}\frac{\partial X^{\alpha}}{\partial x^{\mu}}\right)\mathrm{d}x^{\mu} $$
But since a covector is also a 1-form, this should be recoverable using Cartan's formula:
$$ \mathcal{L}_{X}w = \left(\mathrm{d}\iota_{X} + \iota_{X}\mathrm{d}\right)w $$
where $\mathrm{d}$ is the external derivative and $\iota_{X}$ is the internal produce with respect to $X$.
However, when I expand Cartan's formula, I get:
$$\begin{align} \left(\mathrm{d}\iota_{X} + \iota_{X}\mathrm{d}\right)w_{\mu}\mathrm{d}x^{\mu} &= \left(\frac{\partial X^{\alpha}w_{\alpha}}{\partial x^{\mu}} + X^{\alpha}\frac{\partial w_{\mu}}{\partial x^{\alpha}}\right)\mathrm{d}x^{\mu} \\ &= \left(w_{\alpha}\frac{\partial X^{\alpha}}{\partial x^{\mu}} + X^{\alpha}\frac{\partial w_{\alpha}}{\partial x^{\mu}} + X^{\alpha}\frac{\partial w_{\mu}}{\partial x^{\alpha}}\right)\mathrm{d}x^{\mu} \\ &= \left(X^{\alpha}\frac{\partial w_{\mu}}{\partial x^{\alpha}} + w_{\alpha}\frac{\partial X^{\alpha}}{\partial x^{\mu}}\right)\mathrm{d}x^{\mu} + X^{\alpha}\frac{\partial w_{\alpha}}{\partial x^{\mu}}\mathrm{d}x^{\mu}\\ &= \mathcal{L}_{X}w + X^{\alpha}\mathrm{d}w_{\alpha}\\ \end{align}$$
Most likely I've done something wrong (the extra term comes from the product rule on $\frac{\partial X^{\alpha}w_{\alpha}}{\partial x^{\mu}}$, but I'm not sure this the correct way to treat the gradient of $X^{\alpha}w_{\alpha}$), or there's some reason that $X^{\alpha}\mathrm{d}w_{\alpha}$ vanishes. $X^{\alpha}$ and $w_{\alpha}$ are in general not zero, so that implies the components of $w$ are closed, so:
$$ \mathrm{d}w_{\alpha} = 0 $$
but I can't think of any way to justify this, since in general they're just functions on a manifold and functions aren't necessarily closed (are they?).
You incorrectly evaluated $\iota_X(dw)$. The correct formula for that is $$\sum_{\alpha,\mu} X^\alpha\frac{\partial w_\mu}{\partial x^\alpha} dx^\mu - X^\mu\frac{\partial w_\mu}{\partial x^\alpha} dx^\alpha.$$ Note that the latter terms add up to $-\sum\limits_\mu X^\mu dw_\mu$.