Consider the following theorem and it's proof given by Baby Rudin
I fully understand each step in this proof until the part
"So that $ |s_n - s'_n| \leq \epsilon/2 $ by (26)"
My question is the why of it's inequality. If it comes from (26), then the hypothesis of absolute convergence is superflous, since $ |\sum_{k=n} ^ {m }a_k | \leq \epsilon/2$ would also implies the inequality, right?
In the subtraction $s_n - s'_n$, after canceling terms whose indices in the main sequence are the same, at most finitely many terms survive, and each of the surviving terms is equal to $a_t$ or $-a_t$, for some $t \ge N$.
It follows, by $(26)$, that $$-\epsilon \le s_n - s'_n \le \epsilon$$ hence $|s_n - s'_n| \le \epsilon$.
As regards your question as to why, assuming only convergence, but not absolute convergence, the inequality $$\left|\sum_{k=n} ^ {m }a_k\right| \le \frac{\epsilon}{2}$$ wouldn't imply $|s_n - s'_n| \le \epsilon$, note that $$\sum_{k=n}^{m }a_k$$ might achieve being less than or equal to ${\large{\frac{\epsilon}{2}}}$, in absolute value, by virtue of the signs of the terms (e.g., as in a convergent alternating series), but the surviving terms of $s_n - s'_n$ need not preserve those signs, so you can't make an automatic comparison.