On the coordinates in Guillemin and Pollack: Curvilinear or always Cartesian? In Guillemin and Pollack, the authors always refer to coordinates on, say, a $k$-dimensional manifold. They call these coordinates $\{x_1, \cdots, x_k \}$. Are these always Cartesian coordinates inherited from the standard coordinate embedding space?
How does their theory accommodate curvilinear coordinates? Take $\theta \in [0, 2 \pi)$ on $S^1$, for example. Typically, to parametrize $S^1$ they use a parametrization like, $x \mapsto (x,\sqrt{1-x^2})$, but this utilizes the Cartesian coordinates of $\mathbb{R}^2$, the embedding space of $S^1$.
 A: No. Let $r^1,...,r^k$ denote the standard coordinates on $\mathbb{R}^k$ then if we let $p \in M$ and $(U, \phi) = (U,x^1,...,x^k)$ be a chart on $M$. And so,
$$\phi(p) = (r^1 = x^1(p),..,r^k = x^k(p)); \phi^i = r^i \circ \phi$$
It is the $r^i$'s which specify $\phi(p) \in \mathbb{R}^k$. For instance, knowing $(2,3) \in \mathbb{R}^2$ is ambiguous unless you know how the coordinate system is set up since $(y,x) = (2,3) $ is different from $(x,y) = (2,3)$. You should think of the standard coordinates as indication of how far one has to move with respect to each direction to arrive at the point.
If you take $M \subset \mathbb{R}^k$ and choose $r^1,...,r^k$ to be the standard coordinates on $\mathbb{R}^k$ then given any point $p \in M$ you can simply write $p = (r^1(p),...,r^k(p))$. In most books you will see that they take it a bit further and say; consider $\iota: M \to \mathbb{R}^k$ to be the inclusion then set $\bar{r}^i = r \circ \iota$ to specify that you are restricting the coordinates to $M$. 
Edit: Let me be honest, I am no expert in the field at all, and some things I am uncomfortable with; however, this might clear something up for you. Let $(U, \phi) = (U, x^1,x^2,x^3)$ and $(V, \psi) =(V, y^1,y^2,y^3)$ be coordinate charts on $\mathbb{R}^3$ and suppose $U \cap V \not = \emptyset$. If you choose the standard coordinates on $\mathbb{R}^3$ for the first map then;
$$(r^1 = x^1(p), r^2 = x^2(p), r^3 = x^3(p))$$
If you choose $(R, \theta, \phi)$ in the second case then;
$$(R = y^1(p), \theta = y^2(p), \phi = y^3(p)$$
The two charts are connected by the diffeomorphism (overlap map);
$$\psi \circ \phi^{-1}: \mathbb{R}^3 \to \mathbb{R}^3$$
