This is a continuation of this question.
Let $\mathfrak{g}$ be a Lie $R$-algebra with module basis $e_1,\ldots,e_n$ and zero brackets. Then in the Chevalley (co)chain complex, $\Lambda^k\mathfrak{g}$ has basis $\{e_\sigma; \sigma\!\subseteq\![n],|\sigma|\!=\!k\}$ where $e_{\{i_1,\ldots,i_k\}}\!=\!e_{i_1}\!\wedge\!\ldots\!\wedge\!e_{i_k}$ and $(\Lambda^k\mathfrak{g})^\ast$ has dual basis $\{\chi_\sigma; \sigma\!\subseteq\![n],|\sigma|\!=\!k\}$ where $\chi_\sigma(e\tau)={\small\begin{cases}1;\; \sigma=\tau\\0;\;\text{otherwise}\end{cases}}$. All (co)boundary operators are zero, so $H^\ast(\mathfrak{g}\!\times\!\mathfrak{g})$ has basis $\{(\chi_\sigma,\chi_{\sigma'}); \sigma,\sigma'\!\subseteq\![n]\}$.
The multiplication induced by the diagonal map is $\mu^\ast:(\chi_\sigma,\chi_{\sigma'})\mapsto(\chi_\sigma,\chi_{\sigma'})\!\circ\!\mu$ which sends $e_\tau\!\mapsto\!{\small\begin{cases}1;\; \sigma=\tau=\sigma'\\0;\;\text{otherwise}\end{cases}}$. What is the presentation of $H^\ast(\mathfrak{g})$ (as a quotient of $\Lambda_R[x_1,\ldots,x_N]$)?
Write $C(\mathfrak g)$ for the Chevalley-Eilenberg resolution of a Lie algebra $\mathfrak g$. The diagonal map $D:\mathfrak g\to \mathfrak g\times \mathfrak g$ is a Lie algebra map, and one can use is to construct a morphism of complexes $D_*:C(\mathfrak g)\to C(\mathfrak g\times \mathfrak g)$.
On the other hand, there is a map $C(\mathfrak g\times \mathfrak g)\to C(\mathfrak g)\otimes C(\mathfrak g)$. The composition of this with $D_*$ gives a map $$C(\mathfrak g)\to C(\mathfrak g)\otimes C(\mathfrak g).$$ If you apply $\hom_{\mathfrak g}(-,k)$ in all places, this gives a map $$\Delta:C(\mathfrak g)^*\otimes C(\mathfrak g)^*\to C(\mathfrak g)^*$$ which induces the cup product. Notice that my $C(\mathfrak g)^*$ is your $(\Lambda \mathfrak g)^*$.
In concrete terms, the map $\Delta$ is given by $$(\Delta(f,g))(x_1\wedge\cdots\wedge x_{p+q})=\sum_\pi \mathrm{sgn}\pi f(x_{i_1}\wedge\ldots\wedge x_{i_p})g(x_{i_{p+1}}\wedge\ldots\wedge x_{i_{p+q}})$$ when $f$ and $g$ are a $p$-cocycle and a $q$-cocycle, respectively, and the sign in each term comes from the signature of the permutation of the $x_i$s it involves, and the sum is over permutations $\pi=(i_1,\dots,i_{p+q})$ of $(1,\dots,p+q)$ such that $i_1<\cdots<i_p$ and $i_{p+1}<\cdots<i_{p+q}$ (which we call shuffles). To get at this formula you need to make everything above explicit, of course.
Using this formula, it is easy to see that in the abelian case $\chi_\sigma\smile\chi_{\sigma'}$ is zero if $\sigma$ and $\sigma'$ are not disjoint, and $\pm1$ times $\chi_{\sigma\cup\sigma'}$ in case they are disjoint, with the sign coming from a permutation etc. The isomorphism from the cohomology algebra to the exterior algebra is more or less obvious when you write things out.