In the Chevalley-Eilenberg chain complex, what is $ux_i$? What does "trivial $\frak{g}$-module $k$" mean? Below I denote $R=k$ (any commutative unital ring).
How is the augmentation (last map in the resolution) $U(\mathfrak{g})\to R$ defined? As $u\!=\!\sum_w\!r\!_ww \mapsto r_1$ or $u\!=\!\sum_w\!r\!_ww \mapsto \sum_w\!r\!_w$? Is $R$ an $U(\mathfrak{g})$-module via this augmentation? Here $w=x_{i_1}\cdots x_{i_k}$ is a word in $U(\mathfrak{g})=T(\mathfrak{g})/\langle a\otimes b-b\otimes a-[a,b]\rangle=R\langle x_i|x_ix_j-x_jx_i-[x_i,x_j]\rangle$.
In group (co)homology's bar resolution, the augmentation (last map in the resolution) $R[G]\to R$ sends $\sum_g\!r\!_gg \mapsto \sum_g\!r\!_g$ and $R$ is an $R[G]$-module via this map, i.e. $(\sum_g\!r_gg)r=\sum_g\!r_gr$, correct?
I'm using Weibel 6.5.1 and 7.7.1. The statements are not explicit enough.
Why are you thinking that $U(\frak{g}) $ acts on $ \frak{g} $? From the definition of $ d$ , $ u x_i \in U(\frak{g}) $, so it is the opposite $ \frak{g} $ acts on $ U(\frak{g}) $, by adjoint action.
I don't understand your notation $ u\!=\!\sum_w\!r\!_ww $. What are the $ w $ here?
I am not sure about this but, I think the proper map $ U(\frak{g}) \rightarrow $ $k$ is just the projection of the ground field inside $ U(\frak{g}) $. I think this is the only way to get $ d^2 = 0 $.
EDIT: The lie algebra acts on itself by adjoint action, $ g.h = [g,h] $. This extends to a derivation on the tensor algebra, and $ U(\frak{g}) $.
Recall that the universal enveloping algebra has a filtration coming from the grading of the tensor algebra. The zeroth vector space in this filtration is just the ground field.