Solution:

The functional equation has two solutions, $f(x) \equiv 0$ and $f(x) \equiv 2 x$.

Setting $x=0$ and $y=0$ in the functional equation yields $f(f(0))=0$. So there is at least one zero point of $f$. Let $a$ be any of them. Setting $y=a$ gives us $f(2 a)=f(a x)+f(0)$. If $a \neq 0$, then $f$ is a constant function and we know that $f(a)=0$, so it is a zero function, which is indeed a solution.

It remains to investigate the case where $0$ is the only zero point of $f$, i.e. $f(a)=0$ if and only if $a=0$. Furthermore, taking $x=0$ in the functional equation we obtain
$$f(2 y)=f(f(y))$$
If we prove an injectivity of $f$, the previous identity yields $f(y)=2 y$, what is the second solution, as we can easily check.

Now we prove the injectivity of $f$. Firstly, let us examine the set of the fixed points of $f$. This set is non-empty because $0$ is one of its points. Assume that $p$ is any of the fixed points, i.e. $f(p)=p$. Setting $x=-1, y=p$ in the functional equation gives
$$p=f(-f(p)+2 p)=f(-p)-f(p)+f(f(p))=f(-p)$$
Now we set $x=1, y=-p$ in the functional equation and we obtain using proved $f(-p)=p$
$$p=f(f(-p)-2 p)=f(-p)+f(-p)+f(f(-p))=3 p$$
This yields that $p=0$ is the only fixed point of $f$.

Secondly, we choose $x$ so that $x f(y)+2 y=x y$, which yields $x=2 y /(y-f(y))$. This can be done for each $y \neq 0$, since $0$ is the only fixed point of $f$. This substitution gives us
$$f(f(y))=\frac{2 y f(y)}{f(y)-y}=\frac{2 f^{2}(y)}{f(y)-y}-2 f(y)$$
In order to finish the proof of injectivity let us assume that non-zero real numbers $a, b$ satisfy $f(a)=f(b)$. We have already proved that $f(a)=f(b) \neq 0$. The previous identity yields
$$\frac{2 f^{2}(a)}{f(a)-a}-2 f(a)=f(f(a))=f(f(b))=\frac{2 f^{2}(b)}{f(b)-b}-2 f(b)$$
and it follows that $a=b$. The proof of the injectivity is thereby finished.

We set $x=1, y=0$ in the original equation and we obtain
$$f(f(0))=f(0)+f(0)+f(f(0))$$
which implies $f(0)=0$. This identity yields after setting $x=0$ in the original equation
$$f(2 y)=f(f(y))$$
Let us assume that there exists $t_{0} \neq 0$ such that $f\left(t_{0}\right)=0$. For every real number $t$ we obtain by substitution $x=t / t_{0}, y=t_{0}$ in the original equation
$$f\left(2 t_{0}\right)=f(t)+\frac{t}{t_{0}} \cdot f\left(t_{0}\right)+f(f(y))=f(t)+f\left(f\left(t_{0}\right)\right)$$
This equation with (2) gives us $f(t)=0$ for each real number $t$, which is one of the solutions, as we can easily check.

Now we can assume that $f(y) \neq 0$ for every real number $y \neq 0$. For each real number $t$ and $y \neq 0$ we obtain by putting $x=2 t / f(y)$ in the original equation
$$f(2 t+2 y)=f\left(\frac{2 t y}{f(y)}\right)+2 t+f(f(y))$$
We prove that the function $f$ is injective. If $a, b$ are numbers from $\mathbb{R} \backslash\{0\}$ such that $f(a)=f(b)(\neq 0)$ then the substitutions $t=a, y=b$ or $t=b, y=a$ in the previous equation gives
$$\begin{aligned} f(2 a+2 b) & =f\left(\frac{2 a b}{f(b)}\right)+2 a+f(f(b)) \\ f(2 b+2 a) & =f\left(\frac{2 b a}{f(a)}\right)+2 b+f(f(a)) \end{aligned}$$
This two equations directly yields to $a=b$, which proves the injectivity.

The injectivity of the function $f$ together with (2) lead to $f(y)=2 y$ what is the second solution, as we can check.