= Solution
Set $h_n=2^{-n}$ and $t_n=h_n\lceil t/h_n\rceil$. Replace the last sampled value $X_{t_n}$ by $X_t$, and call the resulting absolute-increment sum $W_n(t)$. This is a sum over a genuine <partition of an interval> of $[0,t]$, and
$$
|V_t^{n,|\cdot|}-W_n(t)|\le |X_{t_n}-X_t|\longrightarrow0
$$
by right continuity. This is the <dyadic approximation of total variation>. Let $V(t)$ be the <total variation of a function> $X$ on $[0,t]$, allowing $+\infty$. Every $W_n(t)$ is at most $V(t)$. Conversely, take any finite <partition of an interval> of $[0,t]$. Approximate its interior points from the right by dyadic grid points, retaining the endpoint $t$. For large $n$ these approximating points remain distinct, and right continuity makes their absolute-increment sum converge to that of the original partition. The <triangle inequality> makes $W_n(t)$ at least this selected sum. Taking the <supremum> over all partitions proves
$$
\boxed{\lim_{n\to\infty}V_t^{n,|\cdot|}=V(t),\quad V(t)=\operatorname{Var}_{[0,t]}X\in[0,\infty].}
$$
The limit is the total-variation function; for locally finite variation it is the <total-variation process>.
There is a missing hypothesis in the assertion about being <càdlàg>. A general <càdlàg function> can have infinite <total variation of a function> arbitrarily close to zero. For example,
$$
X_0=0,\qquad X_s=s\sin(1/s)\quad(s>0)
$$
is continuous. At the successive points $s_j=(\pi/2+j\pi)^{-1}$ its values alternate in sign with magnitude $s_j$. The variation along these points dominates a divergent harmonic series. Thus $V(t)=\infty$ for every $t>0$, but $V(0)=0$, so even extended-valued right continuity fails at zero.
The correct <càdlàg> conclusion holds when $X$ has <finite variation> on every compact interval. Indeed $V$ is then finite and nondecreasing, so it has finite left limits. To prove right continuity at $t$, fix $s>t$ and choose a partition of $[t,s]$ whose sum is within $\varepsilon$ of its variation. If $u>t$ lies before the first interior partition point, changing the first endpoint from $t$ to $u$ changes that sum by at most $|X_u-X_t|$. Additivity of <total variation of a function> therefore gives
$$
0\le V(u)-V(t)\le\varepsilon+|X_u-X_t|.
$$
Let $u\downarrow t$ and then $\varepsilon\downarrow0$. This proves right continuity. This proves <càdlàg regularity of finite total variation>. Under that qualification, the jump formula is
$$
\boxed{\Delta V_t=|\Delta X_t|=|X_t-X_{t-}|\qquad(t>0).}
$$
If the variation is infinite, a difference of two infinite variation values is not a defined jump.
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