If $f,g:(X,A)\to (Y,B)$ are homotopic functions under a homotopy $H:X\times [0,1]\to Y$ such that $H(a,t)\in B$ for all $a\in A$ and $t\in [0,1]$, then $f_*=g_*:H_n(X,A)\to H_n(Y,B)$.
This question has already been posted here Homotopic maps induce the same homomorphism for reduced homology groups and here Homotopic maps $(X, A) \to (Y, B)$ induce homotopic maps $X/A \to X/B$ but I still have many doubts, I would like to know where we are using that $f,g:(X,A)\to (Y,B)$ are homotopic in Homotopic maps $(X, A) \to (Y, B)$ induce homotopic maps $X/A \to X/B$.
I think it is enough to show that $f,g:X/A\to Y/B$ are homotopic, because in this case we already know that $f_*=g_*:H_n(X/A)\to H_n(Y/B)$ where $H_n(X/A)=H_n(X,A)$ and $H_n(Y/B)=H_n(Y,B)$. Now if $f,g:(X,A)\to (Y,B)$ are homotopic, how do I prove that $f,g:X/A\to Y/B$ are homotopic? Thank you.
Your main goal is to prove that if $f,g:(X,A)\to (Y,B)$ are homotopic functions under a homotopy $H:X\times [0,1]\to Y$ such that $H(a,t)\in B$ for all $a\in A$ and $t\in [0,1]$, then $f_*=g_*:H_n(X,A)\to H_n(Y,B)$.
This is usually done by working on the level of chain complexes. See any book treating singular homology.
Your idea was to reduce the proof for pairs to the "already known absolute case" by using that $H_n(X,A) = H_n(X/A)$. Unfortunately this is not true in general (see freakish's comments).
However, we can answer your final question: If $f,g:(X,A)\to (Y,B)$ are homotopic, how to show that $f,g:X/A\to Y/B$ are homotopic?
Let $H$ be a homotopy as above and let $p : X \to X/A, q : Y \to Y/B$ denote the quotient maps. Then also $p \times id : X \times I \to (X/A) \times I$ is a quotient map. This is a well-known fact from general toplogy. Now consider the map $H' = q \circ H : X \times I \to Y/B$. It has the property that $H'(a,t) = [B]$ for all $a \in A$, where $[B]$ denotes the common equivalence class of all points in $B$. Hence the function $$H'' : (X/A) \times I \to Y/B, H''([x],t) = H'(x,t)$$ is well-defined and satisfies $H'' \circ (p \times id) = H'$. By the universal property of quotient maps this equation implies that $H''$ is continuous. But now you have $H''([x],0) = H'(x,0) = qf(x) = f''([x])$ where $f'' : X/A \to Y/B$ denotes the map induced by $f$. Similarly $H''([x],1) = g''([x])$. This proves $f'' \simeq g''$.