Help understanding Rudin's proof of the chain rule

What Rudin really means is this: define $$u(t)=\cases{ \frac{f(t)-f(x)}{t-x}-f'(x) & if $t \ne x$, \\ 0 & if $t = x$. }$$ for $t$ near $x$. You can see that $u(t) \to 0$ as $t \to x$ by the definition of the derivative of $f$ at $x$. Clearly, $$f(t)-f(x)=(t-x)[f'(x)+u(t)]$$ as well.

As you say the definition of the derivative is

$$f'(x) = \lim_{t\rightarrow x} \frac{ f(t) - f(x)}{t-x}$$

So when $t$ is close to $x$ we know that $f'(x)$ is close to $\frac{f(t) - f(x)}{t-x}$. In fact we can judge how close it is, and say that $$f'(x) = \frac{f(t) - f(x)}{t-x} - u(t)$$ where $u(t)$ is a function that tells us the error in this approximation, or how close this is to the actual derivative.

So think of $u$ and $v$ in that proof as judging the error in approximation of $f'(x)$ here.