Expressing a differentiable map and curve in a parametrization This is a question mainly about notation that I just cannot seem to understand. I'm reading Do Carmo's book "Riemannian Geometry" on page 7. Here is some context:
(Here, $\alpha : (-\varepsilon, \varepsilon) \to M$ is differentiable,  and $f$ is differentiable on $M$, where $M$ is a differentiable manifold. )
"If we choose a parametrization $\textbf{x}: U \to M^n$ at $p = \textbf{x}(0)$, we can express the function $f$ and the curve $\alpha$ in this parametrization by
$$f\circ \textbf{x}(q) = f(x_1,\dots,x_n), \hspace{14 pt} q = (x_1,\dots,x_n) \in U,$$
and 
$$\textbf{x}^{-1} \circ \alpha(t) = (x_1(t), \dots, x_n(t)),$$
respectively. " Then he goes on to say that $\frac{d}{dt} (f \circ \alpha)\big|_0 = \frac{d}{dt}f(x_1(t),\dots, x_n(t))\big|_0$. 
What confuses me is that if $q = (x_1, \dots, x_n) \in U$ is in the domain of $\textbf{x}$, then how can $f$ be evaluated here, since its domain is $M$?  
Are we identifying the point $q = (x_1, \dots, x_n)$ with its image under $\textbf{x}$ and then just calling it $(x_1,\dots,x_n)$ again? If so, what purpose does this serve that outweighs the confusion?
The other expression causes me even more confusion. Should this type of thing be obvious to me?
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Param}{\mathbf{x}}$You're perfectly correct that $(x_{1}, \dots, x_{n})$ is being used to denote coordinates both in $M$ and in $U$ by identifying a point $q$ in $U$ with its image $\Param(q)$ in $M$.
Keep in mind there are similar (but distinct) abuses of notation throughout calculus that are probably second-nature. For example, when we speak of coordinates $(x_{1}, \dots, x_{n})$ and then write $q = (x_{1}, \dots, x_{n})$, we use the single symbol $x_{i}$ to denote both a function $x_{i}:U \to \Reals$ and its value at a general point: $x_{i} = x_{i}(q)$. (If we were more literal-minded, we might instead write $q_{i} = x_{i}(q)$, so that $q = (q_{1}, \dots, q_{n})$.)
Anyway, we might use $(x_{1}, \dots, x_{n})$ to denote the Cartesian coordinate functions on $U \subset \Reals^{n}$ and $(\tilde{x}_{1}, \dots, \tilde{x}_{n})$ the coordinates on $\tilde{U} = \Param(U)$ induced by the parametrization $\Param$. Now we have more honest expressions:
$$
\tilde{x}_{i} = x_{i} \circ \Param^{-1}:\tilde{U} \to \Reals,\qquad
x_{i} = \tilde{x}_{i} \circ \Param:U \to \Reals.
\tag{1}
$$
More generally, if $\tilde{f}:\tilde{U} \to \Reals$ is a function, define $f:U \to \Reals$, the "representation of $\tilde{f}$ with respect to the parametrization $\Param$" by
$$
\tilde{f} = f \circ \Param^{-1},\qquad
f = \tilde{f} \circ \Param.
\tag{2}
$$
(The second is the "honest" version of do Carmo's "$f \circ \Param(q) = f(x_{1}, \dots, x_{n})$".)
Note, however, that if $\tilde{q} = \Param(q) \in \tilde{U}$ is the point with coordinates $(\tilde{x}_{1}, \dots, \tilde{x}_{n})$, then $q$ is the point of $U$ with coordinates $(x_{1}, \dots, x_{n})$ by (1), so by (2),
$$
\tilde{f}(\tilde{q})
  = \tilde{f}\bigl(\Param(q)\bigr)
  =  f(q),\qquad\text{i.e.,}\quad
\tilde{f}(\tilde{x}_{1}, \dots, \tilde{x}_{n}) = f(x_{1}, \dots, x_{n}).
\tag{3}
$$
Practically, we may "pretend" the Cartesian coordinates are functions on $M$, and treat coordinate neighborhoods on $M$ as open subsets of $\Reals^{n}$. We can therefore compute with curves, tangent vectors, etc., on $M$ just as if we were looking at curves, tangent vectors, etc., in $\Reals^{n}$.
For example, let $\Param(r, \theta) = (r\cos\theta, r\sin\theta)$ be the polar coordinate mapping restricted to, say, the open set $U = \{(r, \theta) : 0 < r, |\theta| < \pi\}$, on which $\Param$ is a diffeomorphism.
The functions $f:U \to \Reals$ and $\tilde{f}:\tilde{U} \to \Reals$ defined by $f(r, \theta) = r^{2}$ and $\tilde{f}(x, y) = x^{2} + y^{2}$ satisfy the "honest" relationship $f = \tilde{f} \circ \Param$, which (3) allows us to express as the more familiar form $r^{2} = x^{2} + y^{2}$. If we view $(r, \theta)$ and $(x, y)$ as Cartesian coordinates on "different copies of $\Reals^{2}$" (i.e., on $U$ and $\Param(U)$, respectively), the equation $r^{2} = x^{2} + y^{2}$ is nonsensical, since the two sides don't even have the same domain.
Similarly, if $\phi$ is a positive function on $(-\pi, \pi)$, its graph $r = \phi(\theta)$ is contained in $U$, and may be parametrized by $(r, \theta) = \bigl(\phi(t), t\bigr)$. Via $\Param$, we identify the path $(x, y) = \alpha(t) = \bigl(\phi(t)\cos t, \phi(t)\sin t\bigr)$ with the "polar graph" $r = \phi(\theta)$, viewed as a condition on $\tilde{U}$. Here, the "honest" relationship is $(r, \theta) = \Param^{-1} \circ \alpha(t) = \bigl(\phi(t), t\bigr)$.
