According to the Jordan decomposition, a necessary and sufficient condition for a function to have finite variation is that it can be expressed as the difference of two increasing functions.
The book I'm reading then says that the ratio of two functions of finite variation is also of finite variation provided that the modulus of the denominator is larger than a positive constant. How can I use the above fact to prove this? Or is more machinery needed to show this statement?
You don't need the decomposition for this; the result can be proved using the definition of "finite variation". The proof from my answer about absolute continuity applied here with few changes:
Suppose $f$ and $g$ are of finite variation. The vector-valued function $\phi(x)=(f(x),g(x))$ is also of finite variation then, and it takes values in some closed rectangle $R=[a,b]\times [c,d]\subset \mathbb R^2$ with $c>0$. The function $\psi(u,v)= u/v$ is Lipschitz in this rectangle, since its gradient is bounded. It is easy to see from the definition that post-composition with a Lipschitz function preserves finiteness of variation (indeed, the variation is at most multiplied by the Lipschitz constant). Hence, the composition $\psi\circ \phi$ has finite variation, and this is precisely $f/g$.