= Solution
A <distribution function> is nondecreasing and right-continuous. Every nondecreasing function has finite left limits, so $F$ is <càdlàg function>[càdlàg]. Along each partition its increments are nonnegative and telescope, giving
$$
V_F(t)=F(t)-F(0).
$$
For each $n$, the set $A^{(n)}=\{s\in[0,n]:\Delta F(s)\geq2^{-n}\}$ is finite because the sum of its positive jumps is at most $F(n)-F(0)$. Every jump belongs to some $A^{(n)}$, so $A_F=\bigcup_nA^{(n)}$ is countable.
For a finite partition, the identity $y^2-x^2=2y(y-x)-(y-x)^2$ gives
$$
F(t)^2-F(0)^2
=2\sum_kF(t\wedge t_k^n)\bigl(F(t\wedge t_k^n)-F(t\wedge t_{k-1}^n)\bigr)
-\sum_k\bigl(F(t\wedge t_k^n)-F(t\wedge t_{k-1}^n)\bigr)^2.
$$
The first sum tends to the <Lebesgue-Stieltjes integral> $2\int_0^tF\,dF$. In the second, intervals containing no prescribed large jump contribute at most their largest increment times $F(t)-F(0)$; first retain finitely many jumps above a threshold and then let the threshold vanish. The limit is therefore $\sum_{s\in A_F\cap(0,t]}|\Delta F(s)|^2$, proving
$$
F(t)^2=F(0)^2+2\int_0^tF\,dF-
\sum_{s\in A_F\cap(0,t]}|\Delta F(s)|^2.
$$
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