Let $(X,A,B)$ be a cellular triad, therefore $X,A,B$ are CW complexes with $X = A \cup B$ (especially $A,B$ are sub complexes of X$.
How to show that $X/B$ is homeomorphic to $X/A \cap B = A \cup B/A \cap B$?
Ideas: According to Whitehead's theorem it suffices to show that homotopy groups $\pi_i(X/B)$ and $\pi_i(X/A \cap B)$ are isomorphical for each $i$.
But since don't know the concretely CW structure of both quotient spaces I don't know how to argue. Inductively on skeletons?
It is wrong. Take $X = \{ 0, 1, 2\}, A = \{ 0, 1 \}, B = \{ 1, 2 \}$. Perhaps you mean $X / B \approx A / (A \cap B)$?
If you want to check this, you have to understand how the quotient $Y / C$ of a cellular pair $(Y,C)$ gets the structure of a CW complex. It consists of all cells of $Y - C$ plus a separate $0$-cell denoted by $[C]$. Define $(Y/C)^0 = Y^0 - C^0 \cup \{ [C] \}$. We have a canonical quotient map $p_0 : Y^0 \to (Y/C)^0$ mapping $C^0$ to $[C]$. The $1$-cells $e_\alpha^1$ in $Y - C$ have attaching maps $\varphi_\alpha^1 : S^0 \to Y^0$ and we attach them by $p_0 \varphi_\alpha^1$ to $(Y/C)^0$. This produces $(Y/C)^1$ and we get a quotient map $p_1 : Y^1 \to (Y/C)^1$ mapping $C^1$ to $[C]$. Now proceed inductively: The $(n+1)$-cells $e_\alpha^{n+1}$ in $Y - C$ with attaching maps $\varphi_\alpha^{n+1} : S^n \to Y^n$ are attached to $(Y/C)^n$ by $p_n \varphi_\alpha^{n+1}$.