I want to show that if $f$ is of bounded variation in $[a, b]$ then $|f|$ is of bounded variation in $[a, b]$. Suppose that $f$ is of bounded variation then if $P=\{x_{0}, \dots, x_{n} \}$ is a partition of $[a, b]$ note that
$$V_{a}^{b}(f)=\sup_{P} \{V_{P}(f) \}$$ From where $V_{P}(f)$ is the variation of $f$ with respect $P$, and by definition
$$V_{a}^{b}(f)=\sup_{P} \{V_{P}(f) \} \geq \sum_{k=1}^{n}|f(x_{k})-f(x_{k-1})|$$
And by the reverse triangle inequality we have that $$\sum_{k=1}^{n}|f(x_{k})-f(x_{k-1})| \geq \sum_{k=1}^{n} ||f(x_{k})|-|f(x_{k-1})||=V_{P} (|f|) $$
In addition, we now that as $f$ is of bounded variation then $V_{a}^{b}(f) < \infty$ should be able to reach that
$$V_{a}^{b}(|f|)=\sup_{P} \{V_{P}(|f|) \} < \infty$$
But I don't see it clearly, any suggestions?
In the second-to-last line you proved that $V_P(|f|) \leq V_P(f)$ for each $P$, so all you have to do is take suprema over partitions of both sides of your inequality; that is: $$V_a^b(|f|) = \sup_P V_P(|f|) \leq \sup_P V_P(f) = V_a^b(f) < \infty$$
showing that $|f|$ is of bounded variation.