Why should we define the differential in Weil model as follows? I could understand $\sum_{j,k} c_{jk}^i \theta_j \wedge \theta_k$ plays a role in the formula because it is the dual of the structure map of $\mathfrak{g}$. The rest of formula looks mysterious to me.

Cartan-Weil model for Equivariant Cohomology

We define its Weil algebra by $W^*(\mathfrak{g}^*)=S^*(\mathfrak{g}^*) \otimes \wedge^*(\mathfrak{g}^*)$ there is also a natural differential operator $d_W$ which makes $W*(\mathfrak{g}^*)$ into a complex. We define $d_W$ as follows:

Choose a basis $e_1,...,e_n$ for $\mathfrak{g}$ and let $e^*_1,...e^*_n$ its dual basis in $\mathfrak{g}^*$. Let $\theta_1,...,\theta_n$ be the image of $e^*_1,...e^*_n$ in $\wedge(\mathfrak{g}^*)$ and let $\Omega_1,...,\Omega_n$ be the image of $e^*_1,...e^*_n$ in $S(\mathfrak{g}^*)$. Let $c_{jk}^i$ be the structure constants of $\mathfrak{g}$, that is $[e_j,e_k]=\sum_{i=1}^nc_{jk}^ie_i$. Define $d_W$ by \begin{eqnarray} d_W\theta_i=\Omega_i- \frac{1}{2}\sum_{j,k} c_{jk}^i \theta_j \wedge \theta_k \end{eqnarray} and \begin{eqnarray} d_W\Omega_i=\sum_{j,k}c_{jk}^i\theta_j \Omega_k \end{eqnarray} and extending $d_W$ to $W(\mathfrak{g})$ as a derivation.