What is the precise definition of 'uniformly differentiable'? Since I'm not familiar with manifold concept, let's restrict ourselves to functions with real domain.
Let $A\subset \mathbb{R}$ and $f:A\rightarrow \mathbb{R}^k$.
What is '$f$ is uniformly differentiable on $A$' referring to?
 A: Let $f:[a,b] \to \mathbb{R}$.  Differentiability means that the limit 
$$
\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
$$ (with the obvious modifications for $x = a,b$) exists, in which case we denote the limit as $f'(x)$.  This definition can be rephrased as saying that there is a function $f':[a,b] \to \mathbb{R}$ which satisfies 
$$
\lim_{h \to 0} \left |\frac{f(x+h) - f(x) - hf'(x)}{h} \right| = 0.
$$
The uniformity here means that we can approximate uniformly in $x$.  
More precisely, given an $\epsilon > 0$ we may find a $\delta > 0$ so that whenever $0 < |h| < \delta$, then 
$$
\left|\frac{f(x+h) - f(x) - hf'(x)}{h}\right| < \epsilon.
$$
It's easy to show that a differentible function is uniformly differentiable if and only if it's differentiable with a continuous derivative.  I believe this is what Rudin has you prove.
Outside of Rudin's book, I don't know if I've ever heard the term "uniformly differentiable" used exactly, and a quick Google search seems to suggest that the term is primarily connected with that problem.
A: As a matter of fact, the most well-known concepts of differentiability are always pointwise, like (where $f^{-1}(\mathbb R)$ is the domain of  $f$)
$$\forall \varepsilon>0, \exists \delta>0, \forall x \in f^{-1}(\mathbb R): 0<|x-x_0|\le \delta \to \left|\frac{f(x)-f(x_0)}{x-x_0}-L\right| \le \varepsilon$$
or the Cauchy-convergence form
$$\forall x_0,\forall \varepsilon>0, \exists \delta>0, \forall x_1,x_2 : \left.\begin{aligned}0<|x_1-x_0|\le \delta\\ 0<|x_2-x_0|\le \delta\end{aligned}\right\}\to \left|\frac{f(x_2)-f(x_1)}{x_2-x_1}-g(x_0)\right| \le \varepsilon$$
However, just like the concept of pointwise continuity ($\forall x_0 \forall \varepsilon \exists \delta \forall x$) is different from uniform continuity ($\forall \varepsilon \exists \delta \forall x_0 \forall x$), the not-generally-used concept "uniform differentiability" can be defined by erasing $x_0$, as
$$\begin{aligned}
\forall \varepsilon>0, \exists \delta > 0, \forall x_1, x_2 \in f^{-1}(\mathbb R), \\
(0<x_2-x_1\le\delta) \to \forall \xi\in (x_1, x_2)&: \\
\qquad \qquad\left|\frac{f(x_2)-f(x_1)}{x_2-x_1}-g(\xi)\right| &\le \varepsilon
\end{aligned}$$
where I do not use the term $g(x_2)$, but any representers $g(\xi)$ instead, like what Riemann integral uses, to keep the symmetry of the formula.
A page of a Chinese textbook The Straightforward Calculus about uniform differentiability
In fact, it is really compatible with the Lagrange mean value theorem associated with the differentiable-preserving property of close subintervals, which can be written as
$$\forall [u,v] \subset f^{-1}(\mathbb R), \exists \xi \in [u,v]:\frac{f(v)-f(u)}{v-u}=g(\xi)$$
where a quantifier $\forall \xi$ is changed into $\exists \xi$.
