Vector $V^\mu$ vs the Lorentz contravariant components $V^a(x)$ The vector $V^\mu$ was related to the Lorentz contravariant components $V^a(x)$
\begin{equation}
V^a(x)=e^a_\mu(x) V^\mu
\end{equation}
where $V^\mu$ was contravariant under the general coordinate transformation.
(D-branes, Clifford Johnson Page 67)
Notice that $a$ and $\mu$ does not have to have the same dimension, and from the context it seemed that $V^a$ had less degree of freedom?($|\{a\}|\leq|\{\mu\}|$)
This looked like a dimension reduction procedure, but was different from the compatification, i.e. with the compatification procedure, the extra index($\mu=25$) was still there, just not observed under the low energy. But the Lorentz contravariant components eliminated the index completely. The similar thing was the basis from the world sheet to target space, but $V^a(x)$ and $V^\mu$ were obviously both the concept of the target space.
On the other hand, the $V^a(x)$ now dependent on the coordinate $x$, but not for $V^\mu$?(or was it an abbreviation?)
Could you give a hint of how the vector $V^\mu$ and the Lorentz contravariant components $V^a(x)$ was used?
Related:What is the Difference between Lorentz Invariant and Lorentz Covariant?
 A: This is just a choice of basis of vector fields in a smooth manifold really. Let $(M,g)$ be a $D$-dimensional smooth manifold with Lorentzian metric. Let $U\subset M$ be one open set and let $x:U\to \mathbb{R}^D$ one local chart. Inside $U$ we have a basis of vector fields, the coordinate frame $\partial_\mu$. They allow you to expand a general vector field $X:M\to TM$, inside that open set, as $$X|_U = X^\mu \partial_\mu\tag{1}.$$
Now, this basis is not necessarily orthonormal. In fact the inner products between basis vectors gives you the metric components in the associate coframe basis, $$g_{\mu\nu}(x)=g\left(\partial_\mu,\partial_\nu\right),\quad x\in U\tag{2}$$
and it is clear that $g_{\mu\nu}\neq \eta_{\mu\nu}$ in general. We can, however, always find one orthonormal basis locally. Let us suppose we have found such a basis in another open set $U'\subset M$, and let us denote such a basis by $e_a$ where $a=0,\dots, D-1$ identifies the different vectors. Since each $e_a$ is just a vector, in the overlap $U\cap U'$, which we shall assume non-empty, we may expand each $e_a$ the coordinate basis: $$e_a = e_a^\mu \partial_\mu.\tag{3}$$
This defines a $D\times D$ matrix $e_a^\mu$ which contains the components of the vectors making up the orthonormal frame. Such orthonormal frame is what in Physics one often calls a vielbein.
Now consider the vector field $X$ again. Inside $U\cap U'$ we have two bases available: the coordinate basis $\partial_\mu$ and the vielbein $e_a$. We can expand $$X = X^\mu \partial_\mu = X^a e_a.\tag{4}$$
Denoting the inverse matrix of $e_a^\mu$ by $e^a_\mu$ you can easily check that the relation between components (4) implies is $X^a = e^a_\mu X^\mu$. For a general vector field both sets of components will depend on the manifold point you evaluate it, i.e., it will in general be non-constant.
There are two main uses for this:

*

*As usual, working with one orthonormal basis often gives great simplifications. This is something that should be familiar from linear algebra and all of Physics really.


*A deeper reason to introduce this construction is to be able to define the concept of spin structures and introduce spinor fields. It is not possible to give a self-contained discussion of this subject in this answer, but the basic idea is that $(M,g)$ has one natural ${\rm SO}(1,d-1)$ bundle defined over it, the bundle of orthonormal frames. A choice of vielbein $e_a$ on an open subset trivializes this bundle over the open set. If you choose an open covering of $M$ and define vielbeins on all of them, the way the vielbeins are matched on the overlaps basically define the structure of the bundle.
Now depending on the topology of $M$ you may use this construction to build a ${\rm Spin}(1,D-1)$ bundle over $M$ called bundle of spin frames. When this is done, you have one spin structure and you can define spinor fields over $M$. So, in a sense, to be able to introduce spinor fields in a general, possibly curved Lorentzian manifold $(M,g)$, the first step is to introduce vielbeins. A minor comment here is that not all manifolds admit spin structures and that those admiting spin structures might have more than one.
A: The $e^{a}_{\mu}$ objects are called vierbein field (or tetrads, or frame fields) and connect a general frame of coordinates (in which vectors are expressed as $V^{\mu}$) to a local frame of reference. The latter is very often a local inertial frame, but it doesn't need to be one in general.
The article on Frame fields in general relativity should clarify your doubts.
