The answers are 1) yes and 2) no. If $g$ is increasing and $h$ is decreasing (or vise-versa), then $g-h$ is monotone. A non-monotone BV function cannot be represented this way. On the other hand, if $f$ is $BV$ and is represented as $f = g - h$ where $f$ and $g$ are increasing, you can also write $f = (-h) - (-g)$ so that $f$ is a difference of two decreasing functions.