I feel really silly asking this question, but....
I'm trying to prove the identity $$ \mathscr{L}_{fX}\omega \;\; =\;\; f\mathscr{L}_X\omega - \omega(X)df $$
for $f \in C^\infty(M), \; X \in \mathcal{X}(M),$ and $\omega \in \Lambda^1(M)$. My approach is to take a vector field $Y$ on $M$ and exploit the fact that the Lie derivative commutes with contraction: \begin{eqnarray*} \mathscr{L}_{fX}\left [ C(Y\otimes \omega)\right ] & = & C \left [(\mathscr{L}_{fX}Y) \otimes \omega + Y\otimes (\mathscr{L}_{fX}\omega)\right ] \\ fX\left[\omega(Y)\right ] & = & \omega\left ([fX, Y]\right ) + \left (\mathscr{L}_{fX}\omega\right )(Y)\\ \longrightarrow \left (\mathscr{L}_{fX}\omega\right )(Y) & =& fX\left[\omega(Y)\right ] - \omega\left ([fX, Y]\right ). \end{eqnarray*}
I have no trouble showing that $\mathscr{L}_X\omega(Y) = X\omega(Y) - \omega([X,Y])$, and I also can compute
$$ [fX,Y] \;\; = \;\; f[X,Y] - df(Y)X. $$
However, plugging this back into my Lie derivative formula yields: \begin{eqnarray*} \left (\mathscr{L}_{fX}\omega\right )(Y) & =& fX\left[\omega(Y)\right ] - \omega\left ([fX, Y]\right ) \\ & = & fX\left[\omega(Y)\right ] - f\omega([X,Y]) + \omega(X)df(Y) \\ & = & f\left (\mathscr{L}_X\omega\right )(Y) + \omega(X)df(Y). \end{eqnarray*}
Where is the source of my minus-sign error?
Your computation is correct; it's the identity you're trying to prove that's wrong.
You mentioned in a comment that this formula is found in "Morianu and O'Neill" -- I think you must have been referring to Andrei Moroianu's Lectures on Kähler Geometry and Barrett O'Neill's Semi-Riemannian Geometry. I checked both references -- it turns out that O'Neill has a slightly different (and correct) formula, while Moroianu has the same incorrect formula you gave.
I think I figured out why Moroianu made this mistake. It comes down to interpreting O'Neill's formula incorrectly.
Here's what O'Neill actually claimed (Exercise $11$(d) on page $53$): $$ L_{fX} = f\, L_X - \mathscr D_{X\otimes df} \text{ (notation as in Exercise 10).} $$ And Exercise $10$ explains that for an arbitrary $(1,1)$-tensor field $B$, the notation $\mathscr D_B$ refers to the tensor derivation whose action on $0$-tensors is trivial and whose action on vector fields is $Y\mapsto B(Y)$, where we interpret a $(1,1)$-tensor field as an endomorphism of the tangent bundle.
Now for the $(1,1)$-tensor $B=X\otimes df$, the action of $\mathscr D_B$ on vector fields is $Y \mapsto df(Y) X$. Moroianu seems to have incorrectly (and very understandably, I might add) assumed that its action on $1$-forms is $\omega \mapsto \omega(X) df$. However, since we're extending $\mathscr D_B$ to act on all tensor fields as a derivation, this is incorrect. Notice that since $\mathscr D_B$ acts trivially on $0$-tensors, is a derivation with respect to tensor product, and commutes with contractions, we have \begin{align*} 0 &= \mathscr D_B(\omega(Y))\\ &= \mathscr D_B (\text{tr}(\omega\otimes Y)) \\ &= \text{tr}(\mathscr D_B(\omega\otimes Y))\\ &= \text{tr}\big(\mathscr D_B(\omega) \otimes Y + \omega \otimes \mathscr D_B(Y)\big)\\ &= \mathscr D_B(\omega)(Y) + \omega ( B(Y )), \end{align*} from which it follows that $\mathscr D_B(\omega)(Y) = - \omega(B(Y))$.
In particular, this implies $$ \mathscr D_{X\otimes df}(\omega)(Y) = - \omega(\mathscr D_{X\otimes df}(Y) ) = - \omega( df(Y) X ) = - \omega(X) df(Y), $$ or more succinctly $\mathscr D_{X\otimes df}(\omega)= - \omega(X) df$. Combining this with the formula O'Neill stated yields the formula you proved.
(A somewhat simpler way to do the proof, by the way, is to use Cartan's magic formula.)