On an informal explanation of the tangent space to a manifold On Spacetime and Geometry of Sean Carroll pg 17, he states that once a basis is chosen for the tangent space to spacetime at point $p$, say $T_p$, consisting of the vectors $\{\hat{e}_0,\dots,\hat{e}_3\}$ in which $\hat{e}_\mu$ points along $x^\mu$, the vector tangent to the curve defined by $x^\mu(\lambda)$ is $$\vec{v}=\frac{dx^\mu}{d\lambda}\hat{e}_\mu$$Non the less I can think of an inmediate counter example. Lets focus on $\mathbb{R}^2$ and use the unit tangent vectors $\hat{r}$ and $\hat{\theta}$ lying along the polar coordinates $r$ and $\theta$ respectively. Then if we have a particle moving to the curve $r(t)$ and $\theta(t)$ where $t$ stands for time, the velocity vector is given by
$$\vec{v}=\frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}$$We want the velocity vector to be the definition to what is tangent to the particle's trajectory, but it is clear that this vector is not a multiple of what Sean defined to be a vector tangent to the curve, that is, is not a multiple of $$\vec{u}=\frac{dr}{dt}\hat{r}+\frac{d\theta}{dt}\hat{\theta}$$
I noted that this problem is solved if instead we chose the basis $\hat{r}$ and $r\hat{\theta}$ which still lie along the coordinate lines and therefore is a basis which Sean contemplates in his description. Non the less, according to Pavel Grinfeld, to arise to this basis one would have to calculate the derivatives of the position vector $$\hat{e}_\mu=\frac{d\vec{r}}{dx^\mu}$$This method gives the correct answer indeed to our problem in $\mathbb{R}^2$ but is not useful in spacetime since we have no notion of how to calculate derivatives at "eyesight" since we don't have ordinary analytical geometry. Can somebody tell me how to calculate correct basis vectors in an arbitrary manifold such that the first equation of the question holds? (Of course without resorting to simply stating that the basis vectors are $\partial_\mu$ since this gives no new geometric insight) 
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\dd}{\partial}$Carroll's chosen basis consists of coordinate vector fields, a local frame uniquely attached to a coordinate system.
The unit vector field $\hat{\theta}$ is not the gradient of the angle function $\theta$, so $\hat{\theta}$ is not a coordinate vector field in Carroll's sense. As you note, $r\hat{\theta}$ is the gradient of $\theta$. In other words, what Carroll would denote "$\hat{\theta}$", you're denoting "$r\hat{\theta}$".
I don't understand your final paragraph, unfortunately, so the preceding observations may not be helpful to you. In case it helps, it's possible that when physicists say "coordinate vector field", they mean "unit vector field parallel to coordinate curves". If so, this is not the mathematical meaning of the term "coordinate vector field".
Edit: Let $M$ be an $n$-manifold, and $(x^{\mu})$ a system of local coordinates in some non-empty open set $U$. Computationally, one regards $U$ as a subset of $\Reals^{n}$ and $(x^{\mu})$ as Cartesian coordinates on $U$. (Literally, $U \subset M$, and the coordinates define a mapping from $U$ to some open set in $\Reals^{n}$.) In this identification, the coordinate vector fields are the Cartesian vector fields $\Basis_{\mu} = \dd_{\mu}$. The value of the field $\dd_{\mu}$ at a point of $U$ may be viewed as:


*

*The $\mu$th standard basis vector.

*The partial differentiation operator in the $\mu$th coordinate.

*The $\mu$th element of the frame field dual to the coordinate $1$-forms $(dx^{\mu})$.
On a general smooth manifold, "magnitude" has no coordinate-free meaning, and it makes no sense to speak of unit vector fields. The coordinate vector fields satisfy relations such as $\dd_{\mu}(x^{\nu}) = \delta_{\mu}^{\nu}$, but, for example, $\|\dd_{\mu}\|^{2}$ makes no sense.
On a manifold equipped with a pseudo-Riemannian metric $g$, a vector $v$ is unit if $|g(v, v)| = 1$, and the gradient of a smooth function $\phi$ is the vector field $\nabla\phi$ satisfying
$$
d\phi(v) = g(\nabla\phi, v)\qquad\text{for all $v$.}
$$
When speaking of the unit polar coordinate fields in the plane, there's implicit use of the Euclidean metric $g$. If $(r, \theta)$ denote polar coordinates and $\dd_{r} = \nabla r$, $\dd_{\theta} = \nabla\theta$ are the coordinate vector fields, then
$$
\|\dd_{r}\|^{2} = 1,\qquad
\|\dd_{\theta}\|^{2} = r^{2};
$$
that is, $\frac{1}{r} \dd_{\theta}$ ($= \hat{\theta}$) is the unit vector field proportional to the coordinate field $\dd_{\theta}$.
