It's probably far easier (and instructive) to prove the fact directly. (Moreover it's possible for a chain complex to have vanishing homology but not be contractible, consider the chain complex
$$\dots \to 0 \to \mathbb{Z} \xrightarrow{2 \cdot} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$$
but the correct statement is indeed that $f_*$ is an isomorphism in homology iff the mapping cone has vanishing homology groups).
Recall that the mapping cone $Z_*$ is given by $Z_n = D_n \oplus C_{n-1}$ and $$d_Z(y,x) = (d_D(y) + f(x), -d_C(x)).$$
$(\implies)$ Suppose that $f_*$ is an isomorphism in homology. Let $[z] \in H_n(Z)$ be some homology class, represented by a cycle $z \in Z_n$. We want to show that $[z] = 0$, i.e. that $z$ is a boundary.
Write $z = (y,x) \in D_n \oplus X_{n-1}$. We know that $z$ is a cycle, i.e. $$d_Z(z) = 0 = (d(y) + f(x), -d(x)) \implies d(x) = 0 \text{ and } f(x) = d(-y).$$ Thus $x \in X_{n-1}$ is a cycle, and so we can consider $[x] \in H_{n-1}(C)$. But $f(x) = d(-y)$ means that $f(x)$ is a boundary, i.e. $f_*[x] = 0$. Since $f_*$ is an isomorphism, hence injective, and $[x] = 0$ too. So $x = d(x')$ is a boundary for some $x' \in C_n$ (Fact 1).
But now $0 = d(y) + f(x) = d(y) + f(d(x')) = d(y + f(x'))$, and so $y + f(x') \in D_n$ is a cycle, $[y + f(x')] \in H_n(D)$. But $f_*$ is an isomorphism, hence surjective, and thus $[y+f(x')] = f_*[x'']$ for some cycle $x'' \in C_n$, hence $y = f(x''-x')$ (Fact 2).
Combining the two facts, we get:
$$d(0,x''-x') = (f(x''-x'), -d(x''-x') = (y,x) = z$$
as we wanted.
$(\impliedby)$ Now suppose that all the homology groups of the mapping cone vanish, i.e. $H_*(Z) = 0$. We want to show that $f_*$ is an isomorphism in homology.
$f_*$ is injective: Let $[x] \in H_n(C)$ be such that $f_*[x] = 0$, i.e. $dx = 0$ and $f(x) = dy$ for some $y \in D_{n+1}$. We want to show $[x] = 0$. But then $(y,-x) \in Z_{n+1}$ is a cycle: $d(y,-x) = (dy - f(x), -dx) = 0$. Since $H_{n+1}(Z) = 0$, it follows that $(y,-x)$ is a boundary, $(y,-x) = d(y',x') = (dy' + f(x'), -dx')$. In particular $x = dx'$ is a boundary, hence $[x] = 0$ as we wanted.
$f_*$ is surjective: let $[y] \in H_n(D)$ be represented by a cycle $y \in D_n$ ($dy = 0$). Then $(y,0) \in Z_n$ is a cycle, so it's a boundary (because $H_n(Z) = 0$): $(y,0) = d(y',x') = (d(y') + f(x'), -dx')$. So $x' \in C_n$ is a cycle, and $[y] = [d(y') + f(x')] = [f(x')] = f_*[x']$ is in the image of $f_* as we wanted.
I know it looks like a lot of words, but the underlying reasoning is not very complex. It's a simple diagram chase, and it mostly boils down to keeping track of all the definitions and hypotheses correctly. There's only one thing to do at each step, and so we do it (and the mapping cone is designed precisely so that everything works out).
Let $CA = \frac{A \times [0,1]}{A \times \{1\}}$, and let $q:A \times [0,1] \rightarrow CA$ be the quotient map. Consider the space $X \cup CA$ identifying $a \in A$ with $[a,0] \in CA$. Consider the open sets $U := X \cup q(A \times [0,\epsilon))$ and $V := q(A \times (\frac{\epsilon}{2},1])$ for $0 < \epsilon < 1$. Note that $X \cup CA = \text{int}(U) \cup \text{int}(V) = U \cup V$, $U$ is homotopy equivalent to $X$, $V$ is homotopy equivalent to a point (the vertex of $CA$), and $U \cap V$ is homotopy equivalent to $A$. Since homotopy equivalences induce isomorphisms in reduced homology groups for all degrees, the reduced homology groups of a point are all trivial, and $H_n(X,A) = \tilde{H}_n(X,A)$ for all $n$ and nonempty $A$, by the Mayer-Vietoris sequence for reduced homology applied to $X \cup CA$ and $H_n(X,A) \cong \tilde{H}_n(X \cup CA)$ for all $n$, we have the long exact sequence
\begin{align*}
\cdots \rightarrow \tilde{H}_n(A) \rightarrow \tilde{H}_n(X) \rightarrow \tilde{H}_n(X,A) \rightarrow \tilde{H}_{n-1}(A) \rightarrow \cdots \rightarrow \tilde{H}_0(X,A) \rightarrow 0.
\end{align*}
We've now derived the long exact sequence of reduced homology groups of a pair $(X,A)$. It only remains to go further to obtain the long exact sequence of ordinary homology groups.
Best Answer
I think the excision theorem on the triple $(\mbox{cone}(f), CX,\{x_0\})$ should help here. Then, the homology of $(\mbox{cone}(f),x_0)$ is isomorphic to the homology of $(X\times[0,1]\sqcup_f Y, X\times [0,1])$ (here we are gluing the end of the cylinder over $X$ to $Y$ by $(x,1)\sim f(x)$).
The exact sequence of this pair should get you the rest of the way because the inclusion map of $X\times [0,1]$ is essentially just $f$ now, and the two spaces deformation retract each onto $Y$ and $X$ respectively.