Show that $H_n(C_f)=0$ Let $C=\{(C_n,\partial_n)\}_{n\in\mathbb{Z}}$ and $C'=\{(C_n',\partial_n')\}_{n\in\mathbb{Z}}$ be chain complexes of vector spaces. Let $f:C\to C'$ be a chain map. Define a new chain complex $C_f=\{(D_n,\overline{\partial}_n)\}$ as follows:
$D_n=C_{n-1}\oplus C_n'$
$\overline{\partial}_n(x,y)=(-\partial_{n-1}(x),\partial_n'(y)+f_{n-1}(x)).$
I've been proposed the following exercise:

If $f_*:H_n(C)\to H_n(C')$ is an isomorphism, then $H_n(C_f)=0$.

What I've done so far:
Let $[(x,y)]\in H_n(C_f)$. I will try to show that $[(x,y)]=[(0,0)]$ by showing that there exists $(c_1,c_2)\in C_n\oplus C_{n+1}'$ such that $\overline{\partial}_{n+1}(c_1,c_2)=(x,y)$. 
By definition, $\overline{\partial}(x,y)=(-\partial_{n-1}(x),\partial_n'(y)+f_{n-1}(x))=(0,0)$. From this, I get $\partial_{n-1}(x)=0$, so $x\in Z_{n-1}(C)$ ($x$ is a cycle in $C$) and it makes sense to consider its homology class. On the other hand, from the second component I get
$[0]=[-\partial_n'(y)]=[f_{n-1}(x)]=f_*[x].$
Since $f_*$ is isomorphism, this implies $[x]=[0]$ and hence $x\in B_{n-1}(C)$ ($x$ is a boundary), which implies that there exists $c_1\in C_n$ such that $-\partial_n(c_1)=x$. 

Now I've got left to find $c_2\in C_{n+1}'$ such that $y=\partial_{n+1}'(c_2)+f_n(c_1)$. I haven't still used the surjectivity of $f_*$, but I don't know how to use it since in general $y$ does not define a homology class. 
How can I find $c_2$?
 A: What underlies this problem is the long homology exact sequence associated to a short exact sequence of complexes, which I will explain.
As in your set-up, let $f: C\to C'$ be a map of complexes, and let $\text{cone}(f)$ denote the cone of the map, which you already explained how to construct. There is a short exact sequence of complexes
$$\tag 10\to C'\to \text{cone}(f) \to C[-1]\to 0$$
where the maps are the usual inclusion and projection. 
Claim The connecting map $H(C[-1])= H(C)[-1] \to H(C')[-1]$ is the map induced by homology by $f$.
Proof. Suppose that $c\in C[-1]$ is a cycle. You can lift this to $(0,c)$, which has differential $(f(c),0)$. This certainly lifts to $f(c)$ in $C'$, so the snake lemma produces the map $[c]\to [f(c)]$, which is precisely $f_*$.
Now you can look at the LES of $(1)$ to conclude. 
A: You can also solve this "by hand".
Suppose that $(c,c')$ represents a cycle in the cone. You want to find $(b,b')$ such that $d(b,b') = (f(b')-db,db') = (c,c')$.
Since $(c,c')$ is a cycle, $dc = f(c')$ and $dc'=0$. Note then that in particular $c'$ represents a cycle and $f(c')$ represents a boundary. Since $f$ induces an isomorphism in homology, this means that $c'$ must be a boundary, that is, $c' = db''$, so we have solved for the second coordinate. (Almost, in fact!) 
Note, moreover, that $a = c-f(b'')$ is such that $da= dc-f(db'') = dc-f(c')=0$, that is, it is a cycle in $C'$. Since $f_*$ is onto, this means there is a cycle $a'$ and some $b$ such that $f(a')-db = c-f(b'')$.
Finally, note that $c' = d(b''+a')$ since $a'$ is a cycle, and that $f(a'+b'')-db = c$. Thus $d(b,a'+b'') = (c,c')$, as you wanted. 
