Quick Query: Big Oh manipulation in Apostol analytic number theory proof

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There has been discussion of Apostol's proof 6.20 on the nonvanishing of $L(1,\,\,\chi )$ for real nonprincipal $\chi$ elsewhere, but my question focuses on a particular manipulation of big-oh notation I am unsure about.

In the proof, Apostol replaces the term $$\sum\limits_{n\,\, \le \,\sqrt x } {\frac{{\chi (n)}}{{\sqrt n }}} \left( {...\,\, + \,\,O\left( {\sqrt {\frac{n}{x}} } \right)} \right)$$ by $$...\,\,\, + \,\,\,O\left( {\frac{1}{{\sqrt x }}\,\sum\limits_{n\,\, \le \,\,\sqrt x } {\left| {\chi (n)} \right|} } \right)$$

Why has the modulus sign been included?

Also, is it safe to assume that because $$\sum\limits_{n\,\, \le \,\,x} {\chi (n)} \,\, = \,\,O(1)$$ then $$O\left( {\frac{1}{{\sqrt x }}\,\sum\limits_{n\,\, \le \,\,\sqrt x } {\left| {\chi (n)} \right|} } \right)\,\, = \,\,O\left( {\frac{1}{{\sqrt x }}} \right)$$

The full proof can be accessed here, but I assume my specific question on how big-oh notation is being handled with the Dirichlet character is independent of the context!

Many thanks.

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Because you need to apply the triangle inequality to bound $\sum\limits_{n\le x}\frac{\chi(n)}{\sqrt n}$.

$$\left|\sum\limits_{n\le x}\frac{\chi(n)}{\sqrt n}\right| \le \sum_{n\le x}\frac{|\chi(n)|}{\sqrt n}$$