First off set $x=0$, from here we get that: $$2f(f(y))=8y \implies f(f(y))=4y$$We see that $f(f(0))=0$ From here we see that $f$ is a injective function Now we reduced the functional equation into the following: $$f(2x+f(y))=4x+4y$$Now set $x=-y$, we have that $f(-2y+f(y))=0=f(f(0))$ Now by injectivity we have that $$f(y)-2y=f(0) \implies f(y)=2y+f(0)$$So now from $f(f(y))=4y$ we get that $f(0)=0$ From here we have that the only solution is $f(y)=2y$ . . .