= Solution
Since the <Riemannian orthonormal frame> is positively oriented, its dual <coframe> satisfies $\omega_g=\omega_1\wedge\cdots\wedge\omega_n$. The form $\nu$ is the <interior product> $\iota_X\omega_g$. The contraction rule for an <exterior product> gives
$$
\iota_{E_i}(\omega_1\wedge\cdots\wedge\omega_n)
=(-1)^{i-1}\omega_1\wedge\cdots\wedge\widehat{\omega_i}\wedge\cdots\wedge\omega_n
=(-1)^{i-1}\theta_i.
$$
One can also obtain the sign by moving $E_i$ from the first argument into slot $i$ of the alternating form, using $i-1$ transpositions. Linearity in $X=\sum_i f_iE_i$ proves the <interior contraction of the Riemannian volume form> formula:
$$
\boxed{\nu=\sum_{i=1}^n(-1)^{i+1}f_i\theta_i.}
$$
The exponents $i-1$ and $i+1$ have the same parity.
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