In Guillmin & Pollack, we take smooth manifolds to be subsets of Euclidean space.
Take a manifold $X$ in $R^N$. By definition $X$ is a smooth $k$-dimensional manifold if it is locally diffeomorphic to $R^k$, in particular that for $x\in X$, there is open $V$ of $x$ in $X$ which is diffeomorphic with $U$ of $v$, open in $R^k$. Call (the inverse of) this diffeomorphism a parameterization $\phi: U \rightarrow V$, $\phi: v \mapsto x$. In particular $\phi$ is a smooth homeomorphism.
However, this basic definition does not seem to be complete (and this is not clarified in the text), since we have not defined what 'smooth' means on a manifold; indeed, the whole point of this definition is to give differential structure to $X$. So it is not strictly speaking well-defined to say that $\phi$ is a smooth homeomorphism $\phi: U \rightarrow V$.
My understanding of the intended meaning is that $\phi$ 'extends' to a diffeomorphism on some open $V'$ of $x$ with $V\subset V'$, in the sense that $\phi: U \rightarrow V'$ is smooth (in the analytical sense, $C^\infty$ partial derivatives) and the image of $\phi$ is exactly $V \subset V'$. So in particular the derivative map $d\phi_v$ is an $N \times k$ matrix whose entries are the usual partial derivatives $\frac{\partial \phi_i}{\partial x_j}$ where the $x_j$ are the coordinates relative to the standard basis of $R^k$, and $\phi_i$ are the coordinates of an image point of $\phi$ relative to the standard basis in $R^N$.
In fact, this formulation seems to be equivalent to just considering a parameterization to be the inverse of the smooth extension from $X$ to $R^k$ of $\phi^{-1}$ (as defined in the text). A smooth extension $F: V'\subset R^N \, \rightarrow \, U\subset R^k$ of $\phi^{-1}: V \subset X \, \rightarrow \, U\subset R^k$ is a function of Euclidean space which is smooth in the analytical sense and which satisfies $F(x)=\phi^{-1}(x)$ for $x\in X$.
In any case, we can see that $\phi$ is not a diffeomorphism since its derivative map cannot be an isomorphism. But by a suitable change of basis, we can take an invertible submatrix from $d\phi_v$ which is $k \times k$, and the range of this submatrix is exactly the tangent space $T_x(X)$.
So $d\phi_x$ is an isomorphism 'up to change of basis', $\phi$ is a homeomorphism onto $V$ and a smooth, continuous function $U \rightarrow V'$, and $\phi$ can only be called a diffeomorphism isofaras we take the word 'smooth' in the case of such functions to be the special definition given above.
Is all of the above correct?
As I recall, $G \& P$ defines a map $f: ((x_1,..,x_n),U) \mapsto ((f_1,...,f_m),V)$ to be smooth if each partial is continuous and exist for all orders.
If $X$ is not open, they define $f$ to be smooth at $x \in X$ if there exist a open set $U$ of $x$ such that $F: U \to F(U)$ is smooth where $F|_X = f(X)$ and this is perfectly fine since differentiable functions which agree on a neighborhood, give the same directional derivative.
Hence smoothness is a local condition!