I am confused in showing that the sequence is exact at $A_{n-1}' \oplus B_{n-1}$. Here is part of my argument.
Note that $(\rho_4+f_5)\circ (f_4,-\delta_4)=\rho_4 \circ f_4-f_5 \circ \delta_4=0$ by commutativity of the diagram. This shows that $\text{im}(f_4,-\delta_4) \subset \text{ker}(\rho_4+f_5)$. Conversely, let $(a',b) \in \text{ker}(\rho_4+f_5)$. Then $\rho_4(a')+f_5(b)=0 \implies f_5(b)=-\rho_4(a')$.
I have no idea how to proceed. Can someone give me some hint? Thanks!
It's too awkward to keep track of the subscripts on the morphisms, so I'll write all $\delta_j$ as $\delta$ etc. I also assume that each $f:C_k\to C'_k$ is an isomorphism, so in particular, $f:C_{n-1}\to C'_{n-1}$ is an isomorphism.
You have $f(b)=-\rho(a')$. Then $$0=-\rho(\rho(a'))=-\rho(f(a'))=f(\delta(b)).$$ As $f:C_{n-1}\to C'_{n-1}$ is an isomorphism, $\delta(b)=0$ and by exactness, $b=-\delta(a^*)$ where $a^*\in A_{n-1}$. Then $$\rho(f(a^*)-a')=\rho(f(a^*))-\rho(a')=f(\delta(a^*)+b)=0$$ so $a^*-a'=\rho(c')$ by exactness. Then $c'=f(c^*)$ since $f:C_n\to C_n'$ is an isomorphism, and $a^*-a'=f(\delta(c))$. I would try $-\delta(c)+a^*$ as a candidate for an element of $A_{n-1}$ mapped to $(a',b)$.