As we introduced a cohomology group of a complex $K$ (for example over $\mathbb{Z}$) as the set $Hom(C,\mathbb{Z})$, we also talked about the cup product to give the cohomology group a ring structure.
In particular, the cup product is defined as follows: We define the cup product on the basis elements of the simplex as
$$\cup : C^n(K) \times C^m(K) \to C^{n+m}(K)$$
$$(\varphi \cup \psi)(\sigma_{n+m}) := \varphi(\sigma_{\leq n}) \cdot \psi(\sigma_{\geq n})$$
and then take the linear map corresponding to the image of the basis elements.
I was wondering about the following observation: Since $\varphi \cup \psi, \varphi,\psi$ are linear maps, we get for a basis element $\sigma$
$2(\varphi \cup \psi)(\sigma)=(\varphi \cup \psi)(\sigma)+(\varphi \cup \psi)(\sigma)=(\varphi \cup \psi)( \sigma + \sigma)=\varphi(\sigma_{\leq n}+\sigma_{\leq n}) \cdot \psi(\sigma_{\geq n} + \sigma_{\geq n})=2\varphi(\sigma_{\leq n}) \cdot 2\psi(\sigma_{\geq n})=4(\varphi \cup \psi)(\sigma)$
So the map cannot be linear anymore. Where is my mistake in this observation?
Your third step is incorrect. It is not true that $(\phi \cup \psi)(2\sigma) = \phi(2\sigma_{\le n}) \psi(2\sigma_{\ge n})$.
The problem is that $2\sigma$ is not a basis element. The formula you gave defined $\phi \cup \psi$ on basis elements and extended linearly, so by definition, $(\phi \cup \psi)(2\sigma) = 2(\phi \cup \psi)(\sigma) = 2\phi(\sigma_{\le n}) \psi(\sigma_{\ge n})$.
Your formula would only make sense if $2\sigma$ was a basis element with $(2\sigma)_{\le n} = 2\sigma_{\le n}$ and $(2\sigma)_{\ge n} = 2\sigma_{\ge n}$, but this is simply not true.