$\newcommand{\hom}{\mathsf{Hom}}\newcommand{\ch}{\mathsf{Ch}}$This question was left as an exercise in my class of topology and I am not able to make enough progress in it.
So, I am posting my proof looking for an attempt here.
Definition: $f,g : C\to D$ two morphisms of complexes. We say that f is homotopic to g if there is a family : $(h_i : C_i \to D_{i+1})_{(i \in \mathbb{Z})}$ of R-linear maps such that : $f_i - g_i = d \circ h_i + h_{i-1} \circ d $ for every $i \in \mathbb{Z}$.
The family $(h_i)_{i\in \mathbb{Z}}$ is called homotopy between $f$ and $g$.
Then show that (1)$\simeq$ is an equivalence relation on the set of $\hom_{\ch(R)}(C,D)$.
(2) If $f_1 \simeq f_2$ and $g_1 \simeq g_2$ then $f_1 \circ g_1 \simeq f_2 \circ g_2$.
Attempt: I have prove reflexivity by using the zero morphism $(O_i)_{i\in\Bbb Z}$.
For symmetry: Let $f\simeq g$ which implies: $f_i -g_i=d \circ h_i + h_{i-1} \circ d$ for every $i\in \mathbb{Z}$.
I have to show $g \simeq f$. If I multiply the above equation by $-1$, I get $g_i -f_i= -d \circ h_{i}- h_{i-1} \circ d$. But how to deal with the $-d$ in 2nd equation so that it comes with $h_i$. Can you please help?
For reflexivity : I have done it because I was given $f_i -g_i=...$ and $g_i -k_i=...$ and I just added them so as to get $f_i -k_i=...$ ,so If $f_i \simeq g_i$ and $g_i \simeq k_i$ , I showed that $f_i \simeq k_i$.
For (2) : I am not able to show this using what is given.
Can you please help me with my questions?
For 2), you're nearly there. Indeed, $g_i-f_i=-d\circ h_i-h_{i-1}\circ d$. Since all maps are homomorphisms, if I define $h'_i:=-h_i$ then I have: $$g_i-f_i=d\circ h'_i+h'_{i-1}\circ d,\,\forall i$$And also $(h'_i)_i$ assembles to a morphism of complexes ($h'_id=dh'_i)$. So we get $g\simeq f$.