Thank you. That solved the problem for me. After a few minutes of verifying the relavant identities, I understand your solution perfectly. But I still don't see what's wrong with my failed attempt. Experience tells me that I'll see it just before I'm done typing this, but if you're reading this I guess it didn't work out that way.
F^*\omega(X_1,\dots,X_k)=\omega(F_*X_1,\dots,F_*X_k)=\sum_I{}^{'} \omega_I\ dy^{i_1}\wedge\dots\wedge dy^{i_k}(F_*X_1,\dots,F_*X_k)
=\sum_I{}^{'} \omega_I\ \sum_J(F_*X_1)(y^{j_1})\cdots (F_*X_1)(y^{j_k})\ dy^{i_1}\wedge\dots\wedge dy^{i_k}\left(\frac{\partial}{\partial y^{j_1}},\dots,\frac{\partial}{\partial y^{j_k}}\right)
Note that
(F_*X_1)(y^{j_1})=X_1(y^{j_1}\circ F)
and that this is a component of X_1 in the coordinate system y\circ F. We also have
dy^{i_1}\wedge\dots\wedge dy^{i_k}\left(\frac{\partial}{\partial y^{j_1}},\dots,\frac{\partial}{\partial y^{j_k}}\right)=\delta^I_J
where the delta is =0 unless there's exactly one permutation P that takes I to J, and =sgn P if there is. Since y\circ F is a coordinate system too, we can also write
d(y^{i_1}\circ F)\wedge\dots\wedge d(y^{i_k}\circ F)\left(\frac{\partial}{\partial (y\circ F)^{j_1}},\dots,\frac{\partial}{\partial (y\circ F)^{j_k}}\right)=\delta^I_J
So the original expression can be written as
\sum_I{}^{'} \omega_I\ \sum_J X_1(y^{j_1}\circ F)\cdots X_1(y^{j_k}\circ F)\ d(y^{i_1}\circ F)\wedge\dots\wedge d(y^{i_k}\circ F)\left(\frac{\partial}{\partial (y\circ F)^{j_1}},\dots,\frac{\partial}{\partial (y\circ F)^{j_k}}\right)
=\sum_I{}^{'} \omega_I\ \sum_J d(y^{i_1}\circ F)\wedge\dots\wedge d(y^{i_k}\circ F)(X_1,\dots,X_k)
which is everything we want, except that \omega_I hasn't changed. I still don't see what I've done wrong, so I'll post this so you can point and laugh.
