If $X = (u_1, u_2, \ldots, u_a, w_1, w_2, \ldots, w_c)$ and $Y = (v_1, v_2, \ldots, v_b, w_1, \ldots, w_c)$ with $u$'s, $v$'s and $w$'s disjoint, then
$$s_{\lambda}(u,w) s_{\mu}(v,w) = \left( \sum_{\alpha,\ \beta} c_{\alpha \beta}^{\lambda} s_{\alpha}(u) s_{\beta}(w) \right) \left( \sum_{\gamma,\ \delta} c_{\gamma \delta}^{\mu} s_{\gamma}(v) s_{\delta}(w) \right)$$
$$=\sum_{\alpha,\ \gamma} s_{\alpha}(u) s_{\gamma}(v) \sum_{\beta,\ \delta}
c_{\alpha \beta}^{\lambda} c_{\gamma \delta}^{\mu} s_{\beta}(w) s_{\delta}(w)$$
$$= \sum_{\alpha,\ \gamma,\ \epsilon} s_{\alpha}(u) s_{\gamma}(v) s_{\epsilon}(w) \sum_{\beta,\ \delta} c_{\alpha \beta}^{\lambda} c_{\gamma \delta}^{\mu} c_{\beta \delta}^{\epsilon}.$$

The first equality follows from $s_{\lambda}(u,w) = \sum_{\alpha} s_{\alpha}(u) s_{\lambda/\alpha}(w)$ (obvious from the definition using Young tableaux) and the second bullet point here. The last equality uses the first bullet point at the same link.