Maurer-Cartan form and left-invariant vector fields A left-invariant vector field on a Lie group $G$ satisfies
$$
(L_g)_*X = X
$$
for all $g\in G$. So we have in particular $\omega(X_p)=(L_{g^{-1}})X_p=X_e$ with $\omega$ the Maurer-Cartan form on $G$.
Please have a look at this derivation of the Maurer-Cartan equation. It says that if $X$ and $Y$ are left-invariant vector fields, then they satisfy 
$$\tag{1}X(\omega(Y))=Y(\omega(X))=0$$
Why is that true? I only get
$$
X(\omega(Y))=X(Y)=[X,Y]
$$
so
$$
X(\omega(Y))-Y(\omega(X))=X(Y)-Y(X)=[X,Y]-[Y,X]=2[X,Y]
$$
 A: You should start from the Cartan Formula which is
$$d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])$$
Then since $X(\omega(Y))$ and $Y(\omega(X))$ are $0$ then you have
$$d\omega(X,Y)+\omega([X,Y])=0.$$
In fact $\omega(Y)$ is a constant since $Y$ is left-invariant. If $\omega(Y)$ is a constant then whatever derivativeyou will apply you will obtain $0$. Being more precise if you have a vector field $X$ which at the end of the story is some kind of derivative and therefore you have $$X(\omega(Y))=0$$. 
A: Allow me to write a answer with distinguished notations, as in  wiki there's a little abuse of notation without distinguishing elements in the Lie algebra and left invariant vector fields, which feels like the main source of confusion.
Let's write $X \in \mathfrak{g}$, $\mathfrak{g} = T_eG$ the Lie algebra of $G$, and its corresponding left invariant vector field $\tilde{X} \in \mathfrak{X}(G)^L$ with $L_{g,*}\tilde{X}_{h} = \tilde{X}_{gh}$ for all $g, h \in G$. Also $\omega_g(\tilde{X}_g) = L_{g^{-1},*}\tilde{X}_g=\tilde{X}_e=X$. Then the formulation in wiki should be clearer now:
Let $X, Y \in \mathfrak{g}$ and $\tilde{X}, \tilde{Y} \in \mathfrak{X}(G)^L$. Since $\omega_g(\tilde{X}_g)=X \in \mathfrak{g}$ is a constant function of $g \in G$ in the same fibre $T_eG$, its Lie derivative w.r.t. any vector field $Z \in \mathfrak{X}(G)$ vanishes
$$
Z(\omega(\tilde{X}))_g=\mathcal{L}_Z\omega(\tilde{X})_g =\left.\frac{d}{dt}\right\vert_{t=0}\phi_g(t)^*\omega(\tilde{X}) = \left.\frac{d}{dt}\right\vert_{t=0} X = 0,
$$
where $\phi_g(t)$ is a flow of $Z$ defined in a neighborhood of $g \in G$, and the calculation becomes
\begin{align}
d\omega(\tilde{X}, \tilde{Y}) & =\tilde{X}(\omega(\tilde{Y}))-\tilde{Y}(\omega(\tilde{X}))-\omega([\tilde{X},\tilde{Y}]) \\
& =\tilde{X}(Y)-\tilde{Y}(X)-\omega([\tilde{X},\tilde{Y}]) \\
&= -\omega([\tilde{X},\tilde{Y}]).
\end{align}
By the way, when we write $XY$ we usually mean applying $Y$ first then $X$. Acting on a smooth function $f \in C^\infty(M)$, $XYf = X(Yf)$. More precisely, in a local coordinate $(U,x^1,\cdots,x^n)$, write $X = \sum X^i\partial_i$ and $Y = \sum Y^j\partial_j$, we have $XYf = \sum X^i \partial_i(Y^j\partial_jf) = \sum X^i (\partial_iY^j)\partial_jf + \sum X^i 
 Y^j \partial_i\partial_jf$.
