**Note**: As explained below, there is a clash of nomenclature between what Morita calls a Maurer--Cartan form and what Cartan introduced (which is described in the wikipedia page, say).

First of all there are two Maurer-Cartan forms: left-invariant and right-invariant. They are one-forms with values in the Lie algebra. If we identify the Lie algebra (=left-invariant vector fields) with the tangent space at the identity, then the left-invariant MC form $\omega$ is such that acting on a vector field $\xi$ on $G$ gives for all $g \in G$,
$$ \omega(\xi)_g = (L_g)_*^{-1} \xi_g, $$
where $L_g$ means left multiplication by $g\in G$. There is a also a right-invariant one-form defined similarly but using right multiplication.

Now suppose that $\xi$ is a left-invariant vector field on $G$. This means that
$$\xi_g = (L_g)_* \xi_e,$$
where $\xi_e$ is the value of $\xi$ at the identity $e\in G$. In that case,
$$\omega(\xi)_g = (L_g)_*^{-1} (L_g)_* \xi_e = \xi_e,$$
which is constant, since it does not depend on $g$.

Now, as you point out, if $X$ and $Y$ are left-invariant vector fields, then it is immediate that $\omega$ satisfies the structure equation:
$$d\omega(X,Y) = -\omega([X,Y]).$$

Now choose a basis $(e_i)$ for the Lie algebra, so that we can write $\omega = \sum_i \omega^i e_i$, where the $\omega^i$ are one-forms on $G$. Notice that $\omega(e_i)=e_i$, whence $\omega^j(e_i) = \delta^j_i$.

Applying the structure equation to $X=e_i$ and $Y=e_j$ you see that, on the one hand,
$$d\omega(e_i,e_j)=-\omega([e_i,e_j]) = - [e_i,e_j] = - f_{ij}{}^k e_k,$$
whence
$$d\omega^k(e_i,e_j) = f_{ij}{}^k.$$
But this is precisely the result of applying
$$-\tfrac12 \sum_{i,j} f_{ij}{}^k \omega^i \wedge\omega^j$$
on $e_i$ and $e_j$, hence the identity
$$d\omega^k = -\tfrac12 \sum_{i,j} f_{ij}{}^k \omega^i \wedge\omega^j.$$

To write down explicitly the Maurer-Cartan forms, it is not hard. You have to compute the derivative of $L_g$ in your chosen coordinates. It is particularly easy if the group $G$ is a matrix group, in which case you have $\omega_g = g^{-1}dg$ and again you have to compute this in your favourite coordinates for $G$.

**Added**

I just realised that I forgot to answer the bit about the second form of the structure equation. That equation is usually confusing at first because the notation hides the fact that $[\omega,\omega]$ also involves the wedge product of one-forms. By definition, $[\omega,\omega]$ is the Lie-algebra valued 2-form on $G$ whose value on vector fields $X,Y$ is given by
$$[\omega,\omega](X,Y) = [\omega(X),\omega(Y)] - [\omega(Y),\omega(X)] = 2 [\omega(X),\omega(Y)].$$
If you now take $X=e_i$ and $Y=e_j$, left-invariant vector fields, you see that
$$-\tfrac12 [\omega,\omega](e_i,e_j) = -[e_i,e_j] = -\sum_k f_{ij}{}^k e_k,$$
agreeing again with $d\omega(e_i,e_j)$.

**Further addition**

This is in response to one of Anirbit's comments. In Morita's book *Geometry of Differential Forms*, he calls *any* left-invariant form on $G$ a Maurer--Cartan form. I don't think that this is standard. For me, as my answer above, the Maurer--Cartan form is Lie algebra valued. The two notions of Maurer--Cartan forms can of course be reconciled. Choose a basis $(e_i)$ for $\mathfrak{g}$ and a canonical dual basis $e^i$ for $\mathfrak{g}^*$. Let $\omega^i$ be the left-invariant one-form which agrees with $e^i$ at the identity. Then $\omega = \sum_i \omega^i e_i$ is what I have been calling *the* (left-invariant) Maurer--Cartan form.

While I'm at it, let me explain the nature of my factors of $2$, since that seems also to be in dispute. For me the wedge product is defined as follows $\alpha \wedge \beta := \alpha \otimes \beta - \beta \otimes \alpha$, without a factor of $\frac12$.