In essence, you're utilizing the fact that manifolds come equipped with charts and that maps defined on manifolds have corresponding local expressions obtained by composing with charts (see the very first paragraph in 16.3).
Consider a theorem of the following type:
Let $X,Y$ be smooth manifolds and $f:X\to Y$ a mapping. If $f$ satisfies some hypotheses $H$ then it satisfies some conclusion $C$.
Very often when we "reduce to open sets in $\Bbb{R}^n$" what we're taking advantage of is that the conclusion $C$ is "(local) diffeomorphism invariant", in the sense that one or more of the following is true:
- if $\alpha:X\to X'$ is a diffeomorphism then the same conclusion holds for $f\circ\alpha^{-1}:X'\to Y$
- if $U\subset X$ is open and $\alpha:U\to X'$ is a diffeomorphism then the same conclusion holds for $f\circ \alpha^{-1}:X'\to Y$
- if $\beta:Y\to Y'$ is a diffeomorphism then the same conclusion for $\beta\circ f:X\to Y'$
- if $V\subset Y$ is open and $\beta:V\to Y'$ is a diffeomorphism then the same conclusion holds for $\beta\circ f:X\to Y'$
and due to this, we're using the fact that manifolds come equipped with charts (which are diffeomorphisms), so that by composing $f$ on the domain and target with chart maps, we get a local representative map for $f$ between open subsets of $\Bbb{R}^{n_1}\to\Bbb{R}^{n_2}$, and then we prove the theorem there (and usually it's obvious there by inspection).
For example, $H$ could be an assumption on the specific form of the maps involved and $C$ could refer to smoothness of a map (this is clearly preserved under composition by diffeomorphisms). Another example is that $H$ could be an assumption placed on the tangent mapping $Tf_x$, and $C$ could be something like "is a local diffeomorphism". In this case, it is once again clear that any linear-algebra condition on $Tf_x$ will be preserved because $\alpha,\beta$ being diffeomorphisms means that their derivatives are linear isomorphisms.
Hopefully, the above is a general enough for you to get a rough idea of what's going on. As a specific illustration, let me consider (part of) the next theorem in Dieudonne (16.6.4)
Let $Y,X_1,X_2$ be smooth manifolds and $f_1:Y\to X_1$ and $f_2:Y\to X_2$ be two mappings. Then, the mapping $f=(f_1,f_2):Y\to X_1\times X_2$ is $C^{\infty}$ if and only if $f_1$ and $f_2$ are $C^{\infty}$.
Recall that a mapping is $C^{\infty}$ by definition if when you compose with charts on the domain and on the target, the corresponding map between open subsets of cartesian spaces is $C^{\infty}$. So, consider arbitrary charts $(V_1,\beta_1)$ on $X_1$ and $(V_2,\beta_2)$ on $X_2$ and $(U,\alpha)$ on $Y$ such that $f_1(U)\subset V_1$ and $f_2(U)\subset V_2$. Then the corresponding product chart on $X_1\times X_2$ is given by $(V_1\times V_2, \beta_1\times \beta_2)$. So, we have that
- $f$ is $C^{\infty}$ on $U$ (by definition) if and only if $(\beta_1\times \beta_2)\circ f \circ \alpha^{-1} = (\beta_1\circ f_1\circ \alpha^{-1}, \beta_2\circ f_2\circ\alpha^{-1})$ is $C^{\infty}$ as a mapping between open subsets of cartesian spaces.
- $f_1$ is $C^{\infty}$ on $U$ (by definition) if and only if $\beta_1\circ f_1\circ \alpha^{-1}$ is $C^{\infty}$ as a mapping between open subsets of cartesian spaces
- $f_2$ is $C^{\infty}$ on $U$ (by definition) if and only if $\beta_2\circ f_2\circ \alpha^{-1}$ is $C^{\infty}$ as a mapping between open subsets of cartesian spaces.
So far the only thing we've done is that we have composed with charts on the domain, and charts on the target, and we have specifically used the product charts on the product manifold. Now, all the maps in question are between open subsets of cartesian spaces. This is what is meant by "reducing to open subsets of $\Bbb{R}^{n}$".
For ease of notation, let me define $\tilde{f_1}:=\beta_1\circ f_1\circ\alpha^{-1}$ and $\tilde{f_2}:=\beta_2\circ f_2\circ\alpha^{-1}$, and $\tilde{f}:= (\beta_1\times\beta_2)\circ f\circ \alpha^{-1}=(\tilde{f_1},\tilde{f_2})$. So, now all we have to prove to complete the theorem is show that $\tilde{f}$ is $C^{\infty}$ if and only if $\tilde{f_1}$ and $\tilde{f_2}$ are. But I'm sure you've seen this proven before when studying differential calculus in $\Bbb{R}^n$ (or Banach spaces if you've studied from Dieudonne's first book).
