With the definition, we can see your answer. You need to prove: $\text{Bd}(A \cup B) \subseteq \text{Bd}A \cup \text{Bd}B$. Then it follows that a finite union of rectifiable sets is rectifiable.
I assume when you write $S_1 \bigcup \dots \bigcup S_n$ you mean a finite union. Plus I assume it was a mistake that the number $n$ of sets in the union is the same as the dimension of your space $\mathbb R^n$.
On the other hand, your statement has "countable" in there, too. Did you want that? A countable union of rectifiable sets need not be rectifiable. For example, a singleton $\{r\}$ is rectifiable, but a countable set, dense in the unit ball, is not rectifiable, although it is a countable union of those singletons.
The answer is no with the following counterexample.
Take $A = B =[0,1]$ and let $S \subset [0,1]^2$ be a dense set which intersects every vertical and horizontal line in exactly one point. The contruction of such a set is discussed here
For each $y \in [0,1]$, the set $S_y = \{x\in [0,1]: (x,y) \in S\} = \{x_y\}$ contains a single point, and, hence, is rectifiable with $v(S_y) = \int_{[0,1]} \chi_{S_y} = 0$. Since $v(S_y)$ is identically $0$ for every $y \in [0,1]$ it is Riemann integrable over $[0,1]$.
The indicator function $\chi_S$ has the property that every neighborhood $V$ in $[0,1]^2$ contains a point $x \in V\cap S$ where $\chi_S(x) = 1$ and a point $y \in V\cap S^c$ where $\chi_S(y) = 0$ .
Consequently, $\chi_S$ is not Riemann integrable since the lower and upper integrals are unequal:
$$0=\underline{\int}_{[0,1]} \chi_S \neq \overline{\int}_{[0,1]} \chi_S = 1,$$
and, therefore, $S$ is not rectifiable.
Best Answer
Hint: Apply a change of variables to get an integral over $S$.
Using the theorem instead:
We show that the latter condition holds as follows: first, note that $\operatorname{Bd}(cS) = c\operatorname{Bd}(S)$. Then, use the following.
Claim: If a set $K$ has measure zero, then so does the set $cK$.
Proof: The case where $c = 0$ is trivial. If $c\neq 0$, proceed as follows. Suppose that $\mu(K) = 0$. Let $\epsilon > 0$ be arbitrary. Let $(U_j)_{j=1}^\infty$ be a collection of open sets with $K \subset \bigcup_j U_j$ and $\sum_j \mu(U_j) = \epsilon/c$. We note that $cU_j$ is open for each $j$, $cK \subset \bigcup_j (cU_j)$, and $\sum_j \mu(cU_j) = \sum_j c \cdot \epsilon/c = \epsilon$.
We conclude that $cK$ is indeed of measure $0$.