= Solution
The <chain map> identity $d_Df=fd_C$ gives, for $(x,y)\in M_i$,
$$
d_f^2(x,y)=\left(d_C^2x,
(-1)^{i-1}f(d_Cx)+(-1)^id_Df(x)+d_D^2y\right)=(0,0).
$$
Thus $M$ is a <chain complex>; it is a sign convention for the algebraic <mapping cone>.
Put $E_i=C_{i-1}$ with differential $d_E=d_C$. Inclusion into the second summand and projection onto the first give a <short exact sequence of chain complexes>
$$
0\longrightarrow D\xrightarrow{\iota}M\xrightarrow{\rho}E\longrightarrow0.
$$
Its <connecting homomorphism> is particularly explicit. If $x\in C_{i-1}$ is a <chain cycle>, lift it to $(x,0)\in M_i$ and take its boundary. The second component is $(-1)^if(x)$, so the <long exact sequence in homology> contains
$$
\cdots\longrightarrow H_i(C)\xrightarrow{(-1)^{i+1}f_*}H_i(D)
\longrightarrow H_i(M)\longrightarrow H_{i-1}(C)
\xrightarrow{(-1)^if_*}H_{i-1}(D)\longrightarrow\cdots.
$$
If every $H_i(M)$ vanishes, the <exact sequence> makes every $f_*$ both injective and surjective. Conversely, if every $f_*$ is an <isomorphism>, the map $H_i(D)\to H_i(M)$ has zero image, while $H_i(M)\to H_{i-1}(C)$ has zero image and zero kernel; hence $H_i(M)=0$.
One can also see the converse directly on <chain cycles>. If $(x,y)\in M_i$ is a <chain cycle>, then $d_Cx=0$ and $d_Dy=-(-1)^if(x)$. Injectivity of the <induced map on homology> $f_*$ gives $x=d_Cz$. Subtracting the <chain boundary> $d_f(z,0)$ leaves $(0,t)$ with $d_Dt=0$. Surjectivity of the <induced map on homology> $f_*$ lets us choose a <chain cycle> $w\in C_i$ and $v\in D_{i+1}$ such that
$$
t=(-1)^{i+1}f(w)+d_Dv.
$$
Then $(0,t)=d_f(w,v)$. Thus every <chain cycle> in $M$ is a <chain boundary>, proving the <mapping cone acyclicity criterion>:
$$
\boxed{H_*(M)=0\quad\Longleftrightarrow\quad f\text{ is a quasi-isomorphism}.}
$$
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