I see this is an old question (possibly a homework), but I found it while googling something else, so let me write down an answer, since I think the comments above are not 100% correct. If you have a chain homotopy between morphisms of complexes $f, f'\colon C_\bullet \to D_\bullet$, then you can use this chain homotopy to construct by hand a (noncanonical) isomorphism of complexes $Cone (f) \cong Cone (f')$.
A chain homotopy between $f$ and $f'$ is a family of morphisms $h_n\colon C_n\to D_{n+1}$ such that
$$\tag{*} f_n - f_n' = d_{n+1}^D \circ h_n + h_{n-1} \circ d_n^C.$$
Now $Cone (f)$ and $Cone (f')$ are built of objects $D_n \oplus C_{n-1}$ and differentials
\begin{align*}
d_n &= \begin{pmatrix}
d_n^D & f_{n-1} \\
0 & -d_{n-1}^C
\end{pmatrix}
: D_n \oplus C_{n-1} \to D_{n-1} \oplus C_{n-2}
, \\
\quad d_n' &= \begin{pmatrix}
d_n^D & f_{n-1}' \\
0 & -d_{n-1}^C
\end{pmatrix}
: D_n \oplus C_{n-1} \to D_{n-1} \oplus C_{n-2} ,
\end{align*}
where we are using the matrix notation for linear maps between direct sums (i.e., if $p : X \to Z$, $q : X \to W$, $r : Y \to Z$ and $s : Y \to W$ are any four linear maps, then $\begin{pmatrix} p & q \\ r & s \end{pmatrix}$ denotes the linear map from $X \oplus Y$ to $Z \oplus W$ that acts as $p + q$ on $X$ and acts as $r + s$ on $Y$).
We can define a morphism
$$u_n = \begin{pmatrix}
id_{D_n} & h_{n-1} \\
0 & id_{C_{n-1}}
\end{pmatrix}\colon Cone (f) \to Cone (f')$$
It's easy to check that it is a morphism of complexes (multiply the matrices and use the identity (*)). In the other direction, you can define a morphism of complexes
$$v_n = \begin{pmatrix}
id_{D_n} & -h_{n-1} \\
0 & id_{C_{n-1}}
\end{pmatrix}\colon Cone (f') \to Cone (f)$$
Now you just multiply matrices to see that $u_\bullet$ and $v_\bullet$ are mutually inverse maps of complexes, QED.
And a similar thing which is used sometimes: if in general, you have a square of morphisms of complexes
$$\require{AMScd}
\begin{CD}
B_\bullet @>f_\bullet>> C_\bullet\\
@VVu_\bullet V @VVv_\bullet V\\
D_\bullet @>g_\bullet>> E_\bullet
\end{CD}$$
that commutes up to homotopy (i.e. there is some chain homotopy $v_\bullet \circ f_\bullet \simeq g_\bullet \circ u_\bullet$), then you can use this chain homotopy to write down a morphism $Cone (f) \to Cone (g)$ that gives you a commutative diagram (with the two squares commuting strictly)
$$\require{AMScd}
\begin{CD}
0 @>>> C_\bullet @>>> Cone (f) @>>> B_\bullet [-1] @>>> 0\\
@. @VVv_\bullet V @VV V @VVu_\bullet[-1]V \\
0 @>>> E_\bullet @>>> Cone (g) @>>> D_\bullet [-1] @>>> 0\\
\end{CD}$$
The formula for this morphism is something like $\begin{pmatrix}
v_n & h_{n-1} \\
0 & u_{n-1}
\end{pmatrix}$.
Best Answer
Well you're looking for a map $A^{n+1}\oplus B^n\to A^{n+1}\oplus B^n$, and your data is a homotopy $f\to g$, that is, a whole bunch of maps $h_n : A^{n+1}\to B^n$.
So you might try to use that.
How about $K:(a,b) \mapsto (a,b+h(a))$ (forgetting the indices) ?
Then let's see how it interacts with the differentials : $d_f(K(a,b)) = d_f(a,b+h(a)) = (d(a), d(b+h(a))+f(a)) = (d(a), d(b)+dh(a) + f(a))$
Now say you have something like $dh+hd = g-f$, so that $dh+f = g-hd$, you get $(d(a), d(b)+g(a) - hd(a))$
Compare with $K(d_g(a,b)) = K((d(a), d(b)+g(a))) = (d(a), d(b)+g(a)+hd(a))$
That's almost the same thing, up to a sign.
Well this is not a problem : it's just that I most likely messed up some of the sign conventions (for instance, $d$ on $A[1]$ is probably something like $-d_A$ if you pay more attention).
So this is just a rough sketch of what it'll look like, but what I'm saying is that $K$ (or some slight modification of it) will probably be a chain map between $C(g)$ and $C(f)$
Now there will be a similar chain map from $C(f)$ to $C(g)$, and with a bit of luck it'll be easy to show that the two are homotopy inverses to one another.