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 $\frac12$.
Your confusion is revealed in this sentence "Or one defines something called the exterior covariant derivative D (see wiki) and then the curvature is simply the exterior covariant derivative of the connection one-form." This is just not true; one does not take the `exterior covariant derivative of the connection $1$-form' to get the curvature.
Let's be precise: Let $P\to M$ be a principal right $G$-bundle, and let $\omega$ be a $\frak{g}$-valued $1$-form on $P$ that defines a connection on $P$ (I won't repeat the well-known requirements on $\omega$). The curvature $2$-form $\Omega = d\omega +\frac12[\omega,\omega]$ on $P$ is the $2$-form that vanishes if and only if it is possible to find local trivializations $\tau: P_U \to U\times G$ such that $\omega = (\pi_2\circ\tau)^*(\gamma)$ where $\gamma$ is the canonical left-invariant $1$-form on $G$.
Now, given a representation $\rho:G\to \text{GL}(V)$, where $V$ is a vector space, one can define an associated vector bundle $E = P\times_\rho V$. Using $\omega$, it is possible to define an 'exterior covariant derivative operator' $D_\omega:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+1})$ (where $A^p\to M$ is the bundle of alternating (i.e., 'exterior') $p$-forms on $M$). The operator ${D_\omega}^2:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+2})$ then turns out to be linear over the $C^\infty$ alternating forms, so it is determined by its value when $p=0$, i.e., by ${D_\omega}^2:\Gamma(E)\to \Gamma(E\otimes A^{2})$, which can be regarded as an section of $\text{End}(E)\otimes A^2$, i.e., a $2$-form with values in $\text{End}(E)$. The formula for this section, when pulled back to $P$, can now be expressed in terms of $\Omega$ in the usual way. In particular, ${D_\omega}^2$ vanishes identically if $\Omega$ does.
Note that one does not take the 'exterior covariant derivative' of the $1$-form $\omega$ anywhere. Instead, one takes the exterior covariant derivative of the exterior covariant derivative of a section of $E$.
I suspect that what you may be trying to do is interpret $\omega$ as a $1$-form on $P$ with values in the trivial bundle $P\times\frak{g}$ and then say that the curvature is the 'exterior covariant derivative' of $\omega$. However, to make this work, you have to specify a connection on the trivial bundle $P\times\frak{g}$, which amounts to choosing a $1$-form $\eta$ on $P$ that takes values in $\text{End}(\frak{g})$ and setting $D_\eta(s) = ds + \eta\ s$. By setting $\eta = \frac12\text{ad}(\omega)$, one gets $D_\eta\omega = \Omega$ (just by definition), so it is possible to do this, but I don't think that this is that useful an observation, since, after all, you could have taken the connection on the trivial bundle $P\times\frak{g}$ to be $\eta = \frac13\text{ad}(\omega)$ (for example) or even $\eta=0$. What justifies the $\frac12$, other than the desire to get the `right' answer?
Best Answer
The question you are asking is a very basic one in the theory of what Élie Cartan called "the method of the moving frame" (in the original French, "la méthode du repère mobile"), so you should be looking that up. Cartan's basic goal was to understand maps of manifolds into homogeneous spaces, say, $f:M\to G/H$, by associating to each such $f$, in a canonical way, a 'lifting' $F:M\to G$ in such a way that the lifting of $\hat f = g\cdot f$ would be $\hat F = gF$ for all $g\in G$. If one could do such a thing, then one could tell whether two maps $f_1,f_2:M\to G/H$ differed by an action of $G$ by checking whether $F_1^*(\gamma) = F_2^*(\gamma)$, where $\gamma$ is the canonical $\frak{g}$-valued left-invariant $1$-form on $G$.
It turns out that it is not always possible to do this in a uniform way for all smooth maps $f:M\to G/H$ (even in the pointed category, which modifies the problem a little bit, but not by much). However, if one restricts attention to the maps satisfying some appropriate open, generic conditions, then there often is a canonical lifting $F$ for those $f$ belonging to this set of mappings, and it can be characterized exactly by requiring that the $1$-form $\omega_F = F^*(\gamma)$ satisfy some conditions. Working out these conditions in specific cases is what is known as the "method of the moving frame".
There's no point in trying to give an exposition of the theory here because it is covered in many texts and articles, but let me just give one specific example that should be very familiar, the Frenet frame for Euclidean space curves.
Here the group $G$ is the group of Euclidean motions (translations and rotations) of $\mathbb{E}^3$ and $H$ is the subgroup that fixes the origin $0\in\mathbb{E}^3$. The elements of $G$ can be thought of as quadruples $(x,e_1,e_2,e_3)$ where $x\in\mathbb{E}^3$ and $e_1,e_2,e_3$ are an orthonormal basis of $\mathbb{E}^3$.
When $f:\mathbb{R}\to\mathbb{E^3}$ is nondegenerate, i.e., $f'(t)\wedge f''(t)$ is nonvanishing, there is a canonical lifting $F:\mathbb{R}\to G$ given by $$ F(t) = \bigl(f(t),e_1(t),e_2(t),e_3(t)\bigr) $$ that is characterized by conditions on $\omega = F^*(\gamma)$ that are phrased as follows: First, $e_1\cdot df$ is a positive $1$-form while $e_2\cdot df = e_3\cdot df = 0$, and, second, $e_2\cdot de_1$ is a positive $1$-form while $e_3\cdot de_1 = 0$.
These conditions take the more familiar form $$ df = e_1(t)\ v(t)dt,\qquad de_1 = e_2(t)\ \kappa(t)v(t)dt, $$ for some positive functions $v$ and $\kappa$ on $\mathbb{R}$, and they imply $$ de_2 = -e_1(t)\ \kappa(t)v(t)dt + e_3(t)\ \tau(t)v(t)dt, \qquad de_3 = -e_2(t)\ \tau(t)v(t)dt, $$ for some third function $\tau$ on $\mathbb{R}$.
Conversely, any $F:\mathbb{R}\to G$ that satisfies the above conditions on $\omega = F^*(\gamma)$ is the canonical (Frenet) lift of a (unique) nondegenerate $f:\mathbb{R}\to\mathbb{E}^3$.
Without the nondegeneracy condition, the uniqueness fails. Just consider the case in which the image of $f$ is a straight line.
There are similar, but, of course, more elaborate, examples for other homogeneous spaces and higher dimensional $M$, but you should go look at the literature if you are interested in this.
Added in response to request in the comment: There are several excellent sources for the method of the moving frame. I'll just list (alphabetically by author) some of my favorites, which means the ones that I find most felicitous:
Élie Cartan, "La théorie des groupes finis et continus et la géométrie différentielle traitées par la méthode du repère mobile", Paris: Gauthier-Villars, 1937. (His style takes some getting used to, and so many say that Cartan is unreadable, but, once you get used to the way he writes, there's nothing like Cartan for clarity and concision. I certainly have learned more from reading Cartan than from any other source.)
Shiing-shen Chern, W. H. Chen, and K. S. Lam, "Lectures on Differential Geometry", Series on University Mathematics, World Scientific Publishing Company, 1999. (Chern learned from Cartan himself, and was a master at calculation using the method.)
Jeanne Clelland, 1999 MSRI lectures on Lie groups and the method of moving frames, available at http://math.colorado.edu/~jnc/MSRI.html. (A nice, short elementary introduction.)
Mark Green, "The moving frame, differential invariants and rigidity theorems for curves in homogeneous spaces", Duke Math. J. Volume 45, Number 4 (1978), 735-779. (Points out some of the subtleties in the 'method' and that it sometimes has to be supplemented with other techniques.)
Phillip Griffiths, " "On Cartan’s method of Lie groups and moving frames as applied to uniqueness and existence questions in differential geometry", Duke Math. J. 41 (1974): 775–814. (Lots of good applications and calculations.)
Thomas Ivey and J. M. Landsberg, "Cartan for Beginners: Differential Geometry Via Moving Frames and Exterior Differential Systems", Graduate Studies in Mathematics, AMS, 2003. (Also contains related material on how to solve the various PDE problems that show up in the applications of moving frames.)
I think that this is enough to go on. I won't try to go into some modern aspects, such as the work of Peter Olver and his coworkers and students, who have had some success in turning Cartan's method into an algorithm under certain circumstances, or the more recent work of Boris Doubrov and his coworkers on applying Tanaka-type ideas to produce new approaches to the moving frame in certain cases.