Cartan's second structure eqaution states that $$\Omega_i^{j}=\mathrm{d} \omega_{i}^{j}-\omega_{i}^{k} \wedge \omega_{k}^{j}=\frac{1}{2} R_{i k l}^{j} \omega^{k} \wedge \omega^{l} $$
The proof is direct:$$\begin{array}{c} \left(\mathrm{d} \omega_{i}^{j}-\omega_{i}^{h} \wedge \omega_{h}^{j}\right)\left(e_{k}, e_{l}\right)=e_{k}\left(\omega_{i}^{j}\left(e_{l}\right)\right)-e_{l}\left(\omega_{i}^{j}\left(e_{k}\right)\right) \\ -\omega_{i}^{j}\left(\left[e_{k}, e_{l}\right]\right)-\omega_{i}^{h}\left(e_{k}\right) \omega_{h}^{j}\left(e_{l}\right)+\omega_{i}^{h}\left(e_{l}\right) \omega_{h}^{j}\left(e_{k}\right) \\ =e_{k}\left(\Gamma_{i l}^{j}\right)-e_{l}\left(\Gamma_{i k}^{j}\right)-\omega^{h}\left(\left[e_{k}, e_{l}\right]\right) \Gamma_{i h}^{j} \\ -\Gamma_{i k}^{h} \Gamma_{h l}^{j}+\Gamma_{i l}^{h} \Gamma_{h k}^{j} \end{array}$$
And notice that $$\begin{aligned} \mathcal{R}\left(e_{k}, e_{l}\right) e_{i}=& \mathrm{D}_{e_{k}} \mathrm{D}_{e_{l}} e_{i}-\mathrm{D}_{e_{l}} \mathrm{D}_{e_{k}} e_{i}-\mathrm{D}_{\left[e_{k}, e_{l}\right]} e_{i} \\ =& \mathrm{D}_{e_{k}}\left(\Gamma_{i l}^{j} e_{j}\right)-\mathrm{D}_{e_{l}}\left(\Gamma_{i k}^{j} e_{j}\right)-\omega^{h}\left(\left[e_{k}, e_{l}\right]\right) \Gamma_{i h}^{j} e_{j} \\ =&\left(e_{k}\left(\Gamma_{i l}^{j}\right)-e_{l}\left(\Gamma_{i k}^{j}\right)+\Gamma_{i l}^{h} \Gamma_{h k}^{j}-\Gamma_{i k}^{h} \Gamma_{h l}^{j}\right.\\ &\left.-\omega^{h}\left(\left[e_{k}, e_{l}\right]\right) \Gamma_{i h}^{j}\right) e_{j} \\ =&\left(\mathrm{d} \omega_{i}^{j}-\omega_{i}^{h} \wedge \omega_{h}^{j}\right)\left(e_{k}, e_{l}\right) \cdot e_{j} \end{aligned}$$
So we get $$\left(d \omega_{i}^{j}-\omega_{i}^{h} \wedge \omega_{h}^{j}\right)\left(e_{k}, e_{l}\right)=R_{i k l}^{j}$$
But the wedge product of 1-form$$(\omega^k\wedge \omega^l)(e_k,e_l)=\omega^k(e_k)\omega^{l}(e_l)-\omega^k(e_l)\omega^{l}(e_k)=1-\delta^k_l \delta^l_k$$
So where is $\frac{1}{2}$?
Is my wedge product formula wrong ? I found the wedge product formula is defined as $$(\omega \wedge \eta)\left(x_{1}, \ldots, x_{k+m}\right)=\sum_{\sigma \in S h_{k, m}} \operatorname{sgn}(\sigma) \omega\left(x_{\sigma(1)}, \ldots, x_{\sigma(k)}\right) \eta\left(x_{\sigma(k+1)}, \ldots, x_{\sigma(k+m)}\right)$$
First, the indices $k$ and $l$ in the right-hand side of the equation $\mathrm{d} \omega_{i}^{j}-\omega_{i}^{k} \wedge \omega_{k}^{j}=\frac{1}{2} R_{i k l}^{j} \omega^{k} \wedge \omega^{l}$ are just dummy indices. Since we want to feed this with $(e_k,e_l)$, we have to change those dummy indices first, say to $m$ and $n$. So we can rewrite the Cartan's structure equation as $$ \mathrm{d} \omega_{i}^{j}-\omega_{i}^{k} \wedge \omega_{k}^{j}=\frac{1}{2} R_{\,i m n}^{j} \omega^{m} \wedge \omega^{n}. \qquad (\star) $$
To compare with the derivation in the text $$ \left(d \omega_{i}^{j}-\omega_{i}^{h} \wedge \omega_{h}^{j}\right)\left(e_{k}, e_{l}\right)=R_{i k l}^{j}, $$ we feed our equation $(\star)$ with the couple $(e_k,e_l)$. If the derivation is true, the rhs of $(\star)$ should be $$ \frac{1}{2}R^j_{\,imn} \, \omega^m\wedge \omega^n \big (e_k,e_l\big) = R^j_{\,ikl}. $$
Indeed, assuming the text use a convention of wedge product as yours, we have $$ \frac{1}{2}R^j_{\,imn} \, \omega^m\wedge \omega^n \big (e_k,e_l\big) = \frac{1}{2} R^j_{\,imn} (\delta_{k}^m \delta^n_l - \delta^m_l \delta^n_k) = \frac{1}{2}(R^j_{\,ikl} - R^j_{\,ilk}) = R^j_{\,ikl}, $$
where the last equality follows by the (anti)symmetry $R^j_{\, ikl} = -R^j_{\, ilk}$.