There are two problems with your approach.
1) Calculation of the limit $\lim_{h \to 0}\{f(c + h) - f(c)\}/h$
2) not using the fact that $f$ has a relative min/max at $c$.
The first problem seems to be result of some misunderstanding of limit concept (please don't mind, many people do have problems with it in the beginning). If the numerator of a fraction tends to $0$ then it does not necessarily mean that the fraction tends to $0$. We also need to analyze the behavior of denominator (after all a fraction is not just numerator). Had it been the case the denominator tends to some non-zero value then your conclusion that fraction tends to $0$ would have been OK. But when both numerator and denominator of a fraction tend to $0$, then the fraction itself could tend to any value ($\pm \infty$ also) or it may oscillate.
Regarding the proof from Apostol, you need to see how he defines a continuous function $Q(x)$ such that $Q(c) = f'(c)$. He wants to do this only to use the fact that $Q(x)$ will have same sign as $Q(c) = f'(c)$ in a certain neighborhood of $c$. This is a technique and it is not really necessary. Even without creating a new function $Q(x)$ you can proceed as follows:
Suppose $f'(c) > 0$ i.e. $$\lim_{h \to 0}\frac{f(c + h) - f(c)}{h} > 0$$ Now if a limit is positive then it means that the function itself is positive for values of $h$ under consideration (informal common sense, a negative thing can not tend to a positive limit, but this can be justified formally. See update below.) Thus $$\frac{f(c + h) - f(c)}{h} > 0$$ for all values of $h$ satisfying $0 < |h| < \delta$ for some $\delta > 0$.
This means that $f(c + h) - f(c)$ has same sign as that of $h$. In other words if $h > 0$ then $f(c + h) > f(c)$ and if $h < 0$ then $f(c + h) < f(c)$. This means that $f$ has neither a relative minimum nor a relative maximum at $c$. This is exactly what Apostol does through $Q(x)$.
The case when $f'(c) < 0$ can be handled similarly. And thus the only option left is $f'(c) = 0$.
Update: If $\lim_{x \to a}f(x) = A > 0$ then show that the function $f(x)$ is positive in a certain neighborhood of $a$ (except possibly at $a$).
Clearly since $A > 0$ we have $\epsilon = A/2 > 0$. Now there exists a $\delta > 0$ such that $|f(x) - A| < \epsilon$ whenever $0 < |x - a| < \delta$. Thus we have $$A - \epsilon < f(x) < A + \epsilon$$ whenever $0 < |x - a| < \delta$. This clearly means that $$0 < \frac{A}{2} = A - \epsilon < f(x)$$ and thus $f(x)$ is positive for $0 < |x - a| < \delta$.
So you've stumbled upon the concept of a limit and how it can be different than evaluating an expression at the value you are approaching.
When we are asking, for example, what the limit of $2 \cdot\frac{x-1}{x-1}$ is as $x$ approaches $1$, what we aren't asking for is what the value of the expression $2 \cdot\frac{x-1}{x-1}$ is at $x = 1$. This expression is undefined at $x = 1$ since you get $\frac{0}{0}$.
But what we want to know is: as you choose $x$ closer and closer to the value $1$, are the values of $2 \cdot\frac{x-1}{x-1}$ getting closer and closer to some value?
Well, as it turns out in my example, it's easy to see that, yes, the values are getting closer to some value. If $x$ is not equal to $1$, $2 \cdot\frac{x-1}{x-1}$ is equal to $2$. So as $x$ gets closer and closer $1$, the expression $\frac{x-1}{x-1}$ is "getting closer and closer" to $2$, since it's already always $2$ for all $x$, which implies it's always $2$ for all $x$ near $1$ (except of course at $x=1$).
So a limit doesn't care about what happens at the value (e.g., at $x=1$). It cares about what the expression looks like as $x$ gets closer and closer to $1$.
Sometimes, evaluating an expression agrees with its limit. For example, consider the expression $x/2$. What happens as $x$ gets closer and closer to $0$? Well, if $x$ is getting really small, so is $x/2$. So when $x$ is near $0$, $x/2$ is near $0$. So we say $\lim \limits_{x \to 0} x/2 = 0$. But if you evaluate $x/2$ at $x=0$, then we also get $0$. When this happens, we say the expression is continuous at $x = 0$. So $f(x) = x/2$ is continuous at $x = 0$.
In the case of your example, $(k/h) = 128 + 16h$ when $h$ is not zero. So, since the $(k/h)$ equals $128 + 16h$ for all non-zero $h$, then as $h$ gets smaller, $(k/h)$ behaves as $128 + 16h$. But $128 + 16h$ is getting closer and closer to $128$ as $h$ goes to $0$ since $16h$ is getting smaller and smaller. That means $k/h$ is getting closer and closer to $128$. So we write $\lim \limits_{h \to 0} k/h = 128$. Note that this doesn't mean $k/0 = 128$. As I discussed above, we don't care about what happens at the value of $h = 0$. We only care about what happens as $h$ gets closer and closer to $0$. $k/0$ is undefined. But $\lim \limits_{h \to 0} k/h = 128$.
Best Answer
The expression we are interested in is:
$\displaystyle \lim_{x\to a} \frac{f(x)-f(a)}{x-a}$
In a comment you talk about "applying the limit laws first". Well, let's try and see how far we get.
Sure, by continuity, and the limit law that "the difference of two limits is the limit of the difference", we have $\displaystyle \lim_{x\to a} (f(x)-f(a)) =0$.
Even easier, we have $\displaystyle \lim_{x\to a} (x-a) =0$.
But now what? It seems as if you want to apply the limit law that "a quotient of limits is the limit of the quotient", or
Well, if stated like this, this limit law is wrong, and if you look closely in your lecture notes or textbooks, you will see (I hope) that instead, it is only true under a special assumption on $S$. Do you see what is this extra condition?
In other words, thinking you can evaluate $$\displaystyle \lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$ by evaluating $$\displaystyle \frac{\lim_{x\to a} (f(x)-f(a))}{\lim_{x\to a} (x-a)}$$ is exactly the wrong step you are looking for which is "inherently missing some crucial information".
The extra condition on the denominator we would need for that manipulation is exactly not satisfied in your case, and so you cannot apply that "law". The limit of the quotient has to be computed another way, and no contradiction arises.