There are many functions $f$ that satisfy the given condition. One of them is given by the following definition:
$$f(n) = \begin{cases} n & \text{if } n \ge 0 \\ -3n & \text{if } n < 0 \end{cases}$$
which can be verified in the following 3 cases.

$n > 0$: $f(f(n) - 2n) = f(-n) = 3n$ and $2f(n) + n = 2n + n = 3n$.

$n < 0$: $f(f(n) - 2n) = f(-5n) = -5n$ and $2f(n) + n = -5n$.

$n = 0$: We have $f(0) = 0$ and $f(f(n) - 2n) = 2f(n) + n = 0$.