This question is motivated by the fact that most of the books of stochastic calculus always prove the basic results of integration by parts formula and the change of variables formula, and at the same time claim that the results:
General change of variables formula: Let $A$ a function of finite variation and $f \in C^{1}$ then $f(A_{t}) - f(A_{0}) = \int_{0}^ {t} f^{'}(A_{s-}) dA_{s} + \sum_{0 < s \leq t}[f(A_{s}) - f(A_{s-}) - f^{'}(A_{-s})\Delta A_{s}]$
General integration by parts formula: Let $f$ and $g$ real-valued functions of finite variation and measurable with respect to the Borel sigma algebra, then for each $t > 0$ we have $f(t)g(t) = f(0)g(0) + \int_{0}^{t} f_{-}dg + \int_{0}^{t}g_{-} df + \sum_{s \leq t} \Delta f(s) \Delta g(s)$
are immediate results from their analogous basic formulas. I do not know where to find a reference for these results, I would really appreciate any help
My attempt
Let $\Delta = \lbrace (x,y) : 0 < x \leq y \leq t \rbrace $ then
\begin{align} \mu_{f}\otimes \mu_{g} (\Delta) = \int_{(0,t]} \int_{(0,y]} df(x) dg(y) &= \int_{(0,t]}(f(y) -f(0) ) dg(y) \\ &= \int_{(0,t]} f(y)dg(y) - f(0)[g(t) -g(0)] \end{align}
using Fubini's Theorem, we also get
\begin{align} \mu_{f}\otimes \mu_{g} (\Delta) = \int_{(0,t]} \int_{[x,t]} dg(y) df(x) &= \int_{(0,t]}(g(t) - g(-x) ) df(x) \\ &= g(t)[f(t) -f(0)] - \int_{(0,t]}g(-x) df(x) \end{align}
Taking these two equations, we obtain: \begin{align} f(t)g(t) - f(0)g(0) = \int_{(0,t]} f(x)dg(x) + \int_{(0,t]}g(-x) df(x) \end{align}
Using the same method, we obtain:
\begin{align} f(t)g(t) - f(0)g(0) = \int_{(0,t]} f(-x)dg(x) + \int_{(0,t]}g(x) df(x) \end{align}
However, I do not know how to obtain the sum of the jumps. I know that there are some classic results about integration such as the following that appears on Mathematical Analysis from Tom Apostol
where $\alpha $ in this case is a step function with finite jumps
As I proved before, we get \begin{align} f(t)g(t) - f(0)g(0) = \int_{(0,t]} f(x)dg(x) + \int_{(0,t]}g(-x) df(x) \end{align}
We know that $\Delta f$ is measurable, since $\Delta f(s) = f(s) - f(s-) $. Now using the fact that the number of discontinuities is countable; the standard procedure of limit by simple functions; and Theorem 7.11 mentioned before, we get
\begin{align} f(t)g(t) - f(0)g(0) &= \int_{(0,t]} f(x)dg(x) + \int_{(0,t]}g(-x) df(x) \\ &= \int_{(0,t]} (f(x-) + \Delta f(x))dg(x) + \int_{(0,t]}g(-x) df(x) \\ &= \int_{(0,t]} f(x-) dg(x) + \int_{(0,t]} \Delta f(x) dg(x) + \int_{(0,t]}g(-x) df(x) \\ &= \int_{(0,t]} f(x-) dg(x) + \sum_{s \leq t} \Delta f(x) \Delta g(x) + \int_{(0,t]}g(-x) df(x) \\ \end{align}
The General change of variables formula follows using this same method.