I'm trying to do exercise 3.12 of the first chapter in Bourbaki's Lie algebras/groups book, but apparently there is a sign error... somewhere. Let me introduce the objects in question:
First, lets fix a field $k$,$\mathfrak g$ a lie algebra over $k$ and $M$ a $\mathfrak{g}$-module. We define $C^p(\mathfrak{g}, M)$ as the $k$-vector space of alternating $p$-linear mappings of $\mathfrak{g}^p$ into $M$ (that is, $C^p(\mathfrak{g}, M) = \hom_k(\Lambda^p\mathfrak{g}, M)$ ) . This has a canonical $\mathfrak{g}$-module structure induced by the adjoit representation, that is, if I haven't messed this up, if $x, x_i \in g$ and $u\in C^p(\mathfrak{g}, M)$ we have $$ (x.u)(x_1, \dots, x_p) = x \cdot u(x_1, \dots, x_p) - \sum_{i \leq p}u(x_1, \dots, [x,x_i], \dots, x_n). $$
Lets call this action $\theta$, so that $\theta(x)u$ is determined by the expression above. There is also a (degree $-1$) map induced by each $y \in \mathfrak{g}$, namely $i(y): C^{p}(\mathfrak{g}, M) \to C^{p-1}(\mathfrak{g}, M)$ $$ i(y) \cdot u(x_2, \dots, x_p) = u(y, x_2, \dots, x_p). $$
Okay, now Bourbaki claims that $\theta(x)i(y) - i(y)\theta(x) = i[x,y]$, but when I compute I get \begin{align} \theta(x)i(y)u(x_2, \dots, x_p) =& x \cdot u(y, x_2, \dots, x_p) & - u([x,y], x_2, \dots, x_p) &- \sum_{i=2}^p u(y, x_2, \dots, [x,x_i], \dots) \\ i(y)\theta(x)u(x_2, \dots, x_p) =& x \cdot u(y, x_2, \dots, x_p) &&- \sum_{i=2}^p u(y, x_2, \dots, [x,x_i], \dots) \end{align} so that $\theta(x)i(y) - i(y)\theta(x) = -i[x,y]$. What did I do wrong?
Your last two displayed equations are wrong. Let's do $p=2$: let $u \in C^2$ and $y, x_2 \in \mathfrak{g}$. Then $i(y)(u) \in C^1$ is the map $$ i(y)(u): x_2 \mapsto u(y, x_2) $$ So $$\begin{align*} (x \cdot (i(y)(u)))(x_2) &= x(i(y)(u)(x_2)) - (i(y)(u))([x,x_2])\\ &= x\cdot u(y,x_2) - u(y, [x,x_2]) \end{align*} $$ This is $\theta(x) i(y) u$, note that there's no term with an $[x,y]$ commutator whereas you have such a term in your second-to-last displayed equation.
On the other hand, $(i(y)(x\cdot u))(x_2) = (x\cdot u)(y,x_2)$ which is $$ x\cdot u(y,x_2) - u([x,y],x_2) - u(y,[x,x_2]) $$ This is $i(y)(\theta(x)(u))$ at $x_2$. Subtracting this from what we computed before gives $u([x,y],x_2) = (i([x,y])(u))(x_2)$.