I was thinking since $\mathcal{L}_X(d(df))$ = $\mathcal{L}_X(d(\omega))$ = $\mathcal{L}_X(g)$, where g is a (0,2)-tensor, and $d(\mathcal{L}_X(df)) = d(d\mathcal{L}_X(f)) = d(d(X[f])$ which is a two-form, so a (0,2)-tensor.
I don't know a lot about Lie derivatives, but I'm guessing the form stays the same, so are they both two-form and thus (0,2)-tensors?
But then $\mathcal{L}_X(df) = d(\mathcal{L}_Xf)$, so would also $d(\mathcal{L}_X(df)) = \mathcal{L}_X(d(df))$, and these two would actually be the same?
For clarification, $X$ is a vector field, $f$ is a function, and $\mathcal{L}_X$ and $d$ are Lie and exterior derivatives.