$\newcommand{\Hom}{\operatorname{Hom}}$
Suppose as you say that $\varphi,\varphi' : C\to C'$ are chain homotopic through $\Gamma: C\to C'$, meaning that
$\varphi - \varphi' = d_{C'}\Gamma + \Gamma d_C$. First, let me just go over the details of the following claim.
Claim 1. For any other complex $D$, the resulting morphisms
$\varphi\otimes 1,\varphi'\otimes 1 : C\otimes D\to C'\otimes D$ are homotopic via $H = \Gamma\otimes 1$.
Proof. There is a cancellation occurring because of the sign rule:
- The differential of $C\otimes D$ is $\delta = d_C\otimes 1 + 1\otimes d_D$, so we compute
$$(\Gamma \otimes 1)\delta = \Gamma d_C\otimes 1 + \Gamma \otimes d_D$$
- The differential of $C'\otimes D$ is $\delta' = d_{C'}\otimes 1 + 1\otimes d_D$, so we compute
$$\delta(\Gamma \otimes 1) = d_C\Gamma \otimes 1 + (1\otimes d_D)(h\otimes 1) \stackrel{!}=d_{C'}\Gamma \otimes 1 - \Gamma \otimes d_D.$$
This thus gives $\delta H +H\delta = (d_{C'}\Gamma +\Gamma d_C)\otimes 1$, and you can conclude. $\square$
I carried out this computation because something similar will happen when computing with $\Hom$, and in fact here the only degree $+1$ map
$$\Hom(C',D) \longrightarrow \Hom(C,D)$$
you can possibly produce with your information is precomposition with $\Gamma$, usually denoted $\Hom(\Gamma ,1)$ or $\Gamma ^*$ (this is what you're writing with the sharp symbol).
Note! We have that $\Gamma ^*(f) = (-1)^{|f|} f\circ \Gamma $ since $\Gamma$ has degree $-1$ (sign rule!).
Claim 2. Precomposition with $\Gamma$ induces a homotopy
$$\Gamma^* : \Hom(C',D) \longrightarrow \Hom(C,D)$$ between precomposition with $\varphi$ and precomposition with $\varphi'$.
Proof. Again, one has to be careful about cancellations and sign rules.
The differential of $\Hom(C,D)$ will be $\delta(f) = d_Df -(-1)^{|f|} fd_C$. These type of differentials have the property that they behave well for composition: for two composable maps of complexes $g : X\to Y$ and $f: Y\to Z$, and writing $\delta$ for the three possible differentials, we have that
$$\delta(f\circ g) = \delta(f)\circ g + (-1)^{|f|} f\circ \delta g.$$
With this at hand, we compute for a map $f : C' \to D$ that:
$$ \delta(\Gamma^*f) = (-1)^{|f|}\delta(f\circ \Gamma) =
(-1)^{|f|}\delta(f)\circ \Gamma + (-1)^{|f|+|f|}f\circ\delta(\Gamma).$$
This has two summands:
$$\tag{1} (-1)^{|f|}(d_Df -(-1)^{|f|} fd_{C'})\circ \Gamma =
(-1)^{|f|}d_Df\Gamma - fd_{C'}\Gamma$$
$$\tag{2} f\circ\delta(\Gamma) = f\circ (d_{C'}\Gamma + \Gamma d_C).$$
Note this last summand is $f\circ (\varphi-\varphi') =
(\varphi-\varphi')^*(f) = \varphi^*(f) -\varphi'^*(f)$.
At the same time, since $\delta(f)$ has degree $|f|-1$ we compute that
$$\tag{3}
\Gamma^*\delta(f) = (-1)^{|f|-1}\delta(f)\circ \Gamma =
(-1)^{|f|-1}(d_Df-(-1)^{|f|} f\circ d_{C'})\circ \Gamma$$
$$=
-(-1)^{|f|}d_D f \Gamma
+ f d_{C'} \Gamma.$$
Examining thing closely, you can see that $(1)$ and $(3)$ cancel, and all we are left with is the equation
$$\Gamma^*(\delta f)+ \delta \Gamma^*(f) = \varphi^*(f) -($\varphi')^*(f),$$
which says precisely that $\Gamma^*$ is a homotopy between
$ \varphi^*$ and $ (\varphi')^*$. $\square$
Phew! Now you can try and do it for the other coordinate in $\Hom$.
