This is routine (and I am quite sure covered by standard textbooks), although somewhat tedious. First, for a compactly supported, non-negative and continuous $f$, one writes
$$ \tag{1} S_t[f] := \sum_{s \leqslant t} f(s, X_{s-}, X_s) = \lim_{n \to \infty} \sum_{i = 0}^{\lfloor n t\rfloor} f(\tfrac in, X_{(i-1)/n}, X_{i/n}) , $$
which shows that the left-hand side is measurable. Next, for an open and bounded $B$, one approximates the sum of $\mathbb 1_B(s, \Delta X_s)$ by $S_t[f]$ for an appropriate sequence of functions $f$. Finally, one uses the monotone class theorem, or Dynkin's lemma, to show measurability for all Borel sets $B$.

*Remark.* Judging by your previous questions, as well as the emphasized part of this question, you seem to be confused by the use of different $\sigma$-algebras. The above construction does not work in the framework of the product $\sigma$-algebras on the space $\mathbb R^{[0,\infty)}$ of *all* paths, because the set of càdlàg paths is not measurable with respect to the product $\sigma$-algebra. Instead, one works with $D([0, \infty), \mathbb R)$, the class of *càdlàg* paths, with the "cylindrical" $\sigma$-algebra $\mathcal A$ (which is just the previous product $\sigma$-algebra restricted to the non-measurable subset $D([0, \infty), \mathbb R)$). Finally, one augments this $\sigma$-algebra $\mathcal A$ with respect to an appropriate probability measure (or at least one considers the $\sigma$-algebra of universally measurable sets) in order to have various objects (such as hitting times) measurable.

*Edit — a sketch of the proof of (1):* This is a pathwise result.

We may restrict our attention to a finite time horizon: $t \in [0, T]$ for $T$ large enough, so that $f(s, x, y) = 0$ when $s \geqslant T$. Choose $\epsilon > 0$ such that $f(s, x, y) = 0$ if $|x - y| < \epsilon$, and enumerate all jumps of $X_s$ of size larger than $\tfrac\epsilon2$: these times form an increasing sequence $s_k$. Call $X_t'$ the same path as $X_t$, but with all jumps at times $s_k$ removed:
$$ X_t' = X_t - \sum_{k : s_k \leqslant t} \Delta X_{s_k} . $$
Then $|\Delta X_t'| \leqslant \tfrac\epsilon2$ for all $t$.
As in the proof of uniform continuity of continuous functions, one has the following property (see below for a proof): $$ \tag{2} \text{there is $\delta > 0$ such that if $|t_1 - t_2| < \delta$, then $|X_{t_1}' - X_{t_2}'| < \epsilon$.} $$ (Here, of course, $0 \leqslant t_1, t_2 \leqslant T$.)

Suppose that $\tfrac1n<\epsilon$. Then it follows that $f(\tfrac in, X_{(i-1)/n}, X_{i/n}) \ne 0$ only if the interval $[\tfrac{i-1}n, \tfrac in]$ contains some $s_k$ (for otherwise the increments of $X_t$ on this interval are equal to the increments of $X_t'$). The desired result follows now easily by continuity of $f$.

Proof of (2): Suppose, contrary to our claim, that no such $\delta$ exists, and let $p_n, q_n$ satisfy $|p_n - q_n| < \tfrac1n$ and $|X_{p_n}' - X_{q_n}'| \geqslant \epsilon$. By passing to a subsequence, we may assume that $p_n$ and $q_n$ converge to some limit $s$, and it follows that necessarily $|\Delta X_s'| \geqslant \epsilon$.

Limit Theory for Stochastic Processes, 2ed. Springer, 2003" Prop. II.1.16, p.69. This proposition answer your question. $\endgroup$