(i) If $f:A \rightarrow A^{'}$ is a chain map, there is an exact sequence $$0 \rightarrow A^{'} \overset{i}{\rightarrow} M(F) \overset{p}{\rightarrow} A^{+} \rightarrow 0$$ where $i_{n}:A_{n}^{'} \rightarrow A_{n-1}\bigoplus A_{n}^{'}$ is given by $a^{'}_{n}\rightarrow (0,a^{'}_{n})$, and $p_{n} :A_{n-1} \bigoplus A^{'}_{n} \rightarrow A_{n-1}$ is given by $(a_{n-1} , a_{n}^{'}) \rightarrow a_{n-1}$ .
(ii) The corresponding long exact sequence is $$... \rightarrow H_{n+1}(A^{'}) \rightarrow H_{n+1}(M(f)) \rightarrow H_{n}(A)\overset{i_n}{\rightarrow} H_{n}(A^{'}) \rightarrow H_{n}(M(f)) \rightarrow ... $$ (iii) The connecting homomorphism $\partial _{n} :H_{n}(A) \rightarrow H_{n}(A^{'})$ is $f_{*}$, the map induced by $f$.
(iv) $f_{*}$, is an isomorphism for each $n$ if and only if $M(f)$ is acyclic.
for the last statement I will explain a little:
suppose that $f:(A,d) \rightarrow (A^{'},d^{'})$ is a morphism of chain complex,consider
$$...\rightarrow M_{n+1} \overset{\delta_{n+1}}{\rightarrow} M_{n} \overset{\delta_{n}}{\rightarrow} M_{n-1} ...$$
which $$M_{n}=A_{n-1} \oplus A^{'}_n$$
and $\Delta_{n} :M_{n} \rightarrow M_{n-1}$ by $$\Delta_{n}:(a_{n-1},a_{n}^{'}) \rightarrow (-d_{n-1}a_{n-1},d^{'}a^{'}+f_{n-1}a_{n-1}).$$
I must show that this sequence is exact if and only if $$H_{n}(f) :H_{n}(A) \rightarrow H_{n}(A^{'})$$
$$x+Im d_{n+1} \rightarrow f_{n}(x)+Im d^{'}_{n+1}$$
is isomorphism,it will be great if you help me to achieve this.thanks.
new edit:again I put some my new effort for last part(I have solved others and it just remains the last part):
we know that a complex is an exact sequence if and only if all its homology groups is zero.
suppose $M(f)$ is exact sequence then $H_{n+1}(M(f))=0$ and $H_{n}(M(f))=0$.
with cinsidering exact sequence of (ii) we have $H_{n+1}(p)=0$ and $H_{n}(i)=0$ then by exactness of the sequence ,$H_{n}(f)$ will be injective and surjective.
now suppose that $H_{n}(f)$ is isomorphism, we should show that $M(f)$ is exact sequence.and it is sufficient to show that its homology group is zero.
because $H_{n}(f)$ is injective and sequence in (ii) is exact we have $ImH_{n+1}(p)=0$ and then $H_{n+1}(p)=0$. because $H_{n}(f)$ is surjective and sequence in (ii) is exact we have $H_{n}(A^{'})=kerH_{n}(i)$ then $H_{n}(i)=0$,now I want to show that $H_{n+1}(M(f))=0$ and $H_{n}(M(f))=0$ and my problem is in this part,so it will be great if you help me with this.
For a fixed map of complexes $f:A\longrightarrow B$, you already exhibited a short exact sequence $0\to B\to C(f)\to A[-1]\to 0$ so that the connecting morphism $$H_s(A[-1])=H_{s-1}(A) \longrightarrow H_{s-1}(B)$$
is the map induced by $f$ in homology. If follows that if $C(f)$ is acyclic, the long exact sequence breaks up into exact sequences
$$0\to H_{s-1}(A) \longrightarrow H_{s-1}(B)\to 0$$
that show $f$ induces isomorphism on homology. Conversely, if $f$ induces isomorphisms on homology, you obtain from the long exact sequence exact sequences
$$0={\rm im} f_* \to H_s(C(f)) \to \ker f_*=0$$
so that $C(f)$ is acyclic.