If I have two chain complexes $C$ and $D$ and I suppose that there is a homotopy between $\phi, \psi:C \rightarrow D$ (i.e there is a sequence of homomorphisms $(K_n: C_n\rightarrow D_{n+1})$ such that $\phi_n-\psi_n=\partial_{n+1}^2\circ K_n+K_{n-1}\circ\partial_n^1$) and I want to prove that $\phi_{*}=\psi_{*}:H_{*}(C)\rightarrow H_{*}(D)$
For this I take a cycle $x\in C$ (i.e $x\in C_n;\partial_n^1 x=0$) so I have that $\psi_n-\phi_n(x)=\partial_{n+1}^2\circ K_n(x)$ why I can deduce that $[\phi_n(x)]=[\psi_n(x)]$?
As you already figured out, you get for a cycle $x\in \ker\partial_n^1$ $$ \phi_n(x) - \psi_n(x) = \partial_{n+1}^2(K_n(x)) \in \operatorname{im} \partial_{n+1}^2. $$ Since $H_n(D) = \ker \partial_n^2/\operatorname{im} \partial_{n+1}^2$, this yields $$ [\phi_n(x) - \psi_n(x)] = [\partial_{n+1}^2(K_n(x))] = 0, $$ thus $[\phi_n(x)] = [\psi_n(x)]$.