Letting $y = 0$, we see that
$$f(2x + f(0)) = 4x - f(0). \quad (1)$$
Replacing $x$ by $(x - f(0))/2$ above, we get
$$f(x) = 2x + c, \quad (2)$$
where $c = -2f(0) - f(0)$. Thus,
$$f(f(x)) = 2(2x + c) + c = 4x + 3c. \quad (3)$$
Using (2) and (3) in the original functional equation, we get
$$2(2x + 2y + c) + c + (4y + 3c) = 4x + 8y,$$

and so $c = 0$. Thus, $f(x) \equiv 2x$ is the only solution.

Letting $x = y = 0$, we see that $f(f(0)) + f(f(0)) = 0$, thus
$$f(f(0)) = 0. \quad (4)$$
Substituting $x = -\frac{1}{2}f(0)$ and $y = 0$ in the original equation and using (4),
we get $f(0) = -2f(0)$, which implies
$$f(0) = 0. \quad (5)$$
Replacing $x$ by $\frac{1}{2}x$ and letting $y = 0$ in the original equation and using (5),
we get
$$f(x) = 2x.$$
Because $2(2x + 2y) + 2(2y) = 4x + 8y$ this indeed is a solution.