The Einstein 3-forms are given (in four dimensions) by \begin{equation} \gamma^i = -\frac12 \Omega_{jk} \wedge {*}(\sigma^i\wedge\sigma^j\wedge\sigma^k) \end{equation} Recalling that the Hodge dual of $\sigma^i$ satisfies \begin{equation} \sigma^i\wedge{*}\sigma^i = g(\sigma^i,\sigma^i)\,\omega \end{equation} (no sum on $i$), we can write \begin{equation} {*}\sigma^i = \frac{1}{3!}\epsilon^i{}_{jkl}\,\sigma^j\wedge\sigma^k\wedge\sigma^l \label{sigdual} \end{equation} where $\epsilon_{ijkl}$ is the alternating symbol in $4$ dimensions, sometimes called the Levi-Civita symbol, which satisfies \begin{equation} \epsilon_{1234}=1 \end{equation} and is completely antisymmetric in its indices. The factor of $3!$ is needed to compensate for the implicit sum, and the raised index takes care of the factor of $g(\sigma^i,\sigma^i)$. Using the identity \begin{equation} -\epsilon_i{}^{pqr} \epsilon^i{}_{jkl} \sigma^j\wedge\sigma^k\wedge\sigma^l = 3!\,\sigma^p\wedge\sigma^q\wedge\sigma^r \end{equation} leads to \begin{equation} -\epsilon_i{}^{pqr} {*}\sigma^i = \sigma^p\wedge\sigma^q\wedge\sigma^r \end{equation} and taking the Hodge dual of both sides now yields \begin{equation} {*}(\sigma^i\wedge\sigma^j\wedge\sigma^k) = -\epsilon_\ell{}^{ijk}({*}{*}\sigma^\ell) = -\epsilon_\ell{}^{ijk}\sigma^\ell \end{equation} since ${*}{*}=1$ for 1-forms in four dimensions with signature $s=-1$. We have therefore shown that \begin{equation} \gamma^i = \frac12 \epsilon_\ell{}^{ijk} \,\Omega_{jk} \wedge \sigma^\ell \end{equation}

Expanding the curvature 2-forms in terms of the Riemann tensor components, we have \begin{equation} \gamma^i = \frac14 \epsilon_\ell{}^{ijk} \,R_{jkpq}\, \sigma^p\wedge\sigma^q\wedge \sigma^\ell \end{equation} and we can now take the Hodge dual of both sides, obtaining \begin{align} {*}\gamma^i &= \frac14 \epsilon_\ell{}^{ijk} R_{jkpq}\, {*}(\sigma^p\wedge\sigma^q\wedge \sigma^\ell) \nonumber\\ &= -\frac14 \epsilon_\ell{}^{ijk} \epsilon_m{}^{pq\ell} \,R_{jkpq}\, \sigma^m \nonumber\\ &= \frac14 \epsilon_\ell{}^{ijk} \epsilon^{\ell pq}{}_m \,R_{jkpq}\, \sigma^m \end{align} Using the identity \begin{align} -\epsilon_\ell{}^{ijk} \epsilon^{\ell pq}{}_m = & \big( \delta^i{}_p\delta^j{}_q\delta^k{}_m + \delta^j{}_p\delta^k{}_q\delta^i{}_m + \delta^k{}_p\delta^i{}_q\delta^j{}_m \nonumber\\ &- \delta^i{}_p\delta^k{}_q\delta^j{}_m - \delta^j{}_p\delta^i{}_q\delta^k{}_m - \delta^k{}_p\delta^j{}_q\delta^i{}_m \big) \end{align} we have \begin{align} {*}\gamma^i = & -\frac14\big( R_{jm}{}^{ij}\,\sigma^m + R_{jk}{}^{jk}\,\sigma^i + R_{mk}{}^{ki}\,\sigma^m \nonumber\\ &\qquad - R_{mk}{}^{ik}\,\sigma^m - R_{jm}{}^{ji}\,\sigma^m - R_{jk}{}^{kj}\,\sigma^i \big) \nonumber\\ = & R_{pm}{}^{pi}\,\sigma^m - \frac12 R_{jk}{}^{jk}\,\sigma^i \nonumber\\ = & \left( R_m{}^i - \frac12 R \,\delta_m{}^i \right) \sigma^m = G_m{}^i \sigma^m = G^i \end{align} and we have shown that components of the Einstein tensor, defined directly in terms of the curvature 2-forms, do indeed take their standard form, that is, that \begin{equation} {*}\vec\gamma = \GG \end{equation} as claimed.