@user681391
Picture. Difference of $\omega$ evaluated on the two flowlines of $X$ should be thought of as $Y\omega(X)$, difference of $\omega$ evaluated on the two flowlines of $Y$ should be thought of as $X \omega(Y)$ and $\omega$ evaluated on the truncation edge is $\omega([X, Y])$. Modulo signs, if you add them up, you get the formula for $d\omega(X, Y)$