Answer: $f(x) = 0$ for all $x \in \mathbb{R}$.
Consider several preliminary substitutions to the problem condition
$$f(f(x) + f(y)) = (x + y)f(x + y). \qquad (1)$$
Substitution $x = y = 0$ gives the equality $f(2f(0)) = 0$. Substitution $x = 2f(0)$, $y = 0$ gives $f(f(0)) = 0$. Finally, substitution $x = y = f(0)$ gives $f(0) = 0$.
Substitution $y = 0$ to (1) implies that for all $x \in \mathbb{R}$ holds $f(f(x)) = xf(x)$. Suppose that for some $a, b \in \mathbb{R}$ holds $f(a) = f(b) \ne 0$, then $af(a) = f(f(a)) = f(f(b)) = bf(b)$, whence $a = b$. Therefore $f$ can attend several times only zero.
We will show that there exists $c \ne 0$ such that $f(c) = 0$. Substitution $y = 1-x$ to (1) implies $f(f(x)+f(1-x)) = f(1)$. If $f(1) = 0$, then $c = 1$. Otherwise, $f(x)+f(1-x) = 1$ for all $x \in \mathbb{R}$. Substitutions of $x = 0$ and $x = 1/2$ in the latter equation give $f(1) = 1$ and $f(1/2) = 1/2$. Finally for $x = 1$ and $y = 1/2$ the equality (1) gives $f(3/2) = 3/2f(3/2)$, whence $f(3/2) = 0$.
Denote by $c$ an arbitrary nonzero number such that $f(c) = 0$. Substitution $x = y = c$ to (1) implies $f(2c) = 0$. Suppose that there exist a real number $t$ such that $f(t) \ne 0$. Putting to (1) successively $x = c$, $y = -c + t$ and $x = 2c$, $y = -2c + t$, we obtain
$$f(f(-c + t)) = tf(t) = f(f(-2c + t)), \text{ where } tf(t) \ne 0.$$
Therefore, $f(-c+t) = f(-2c+t) \ne 0$, whence $-c+t = -2c+t$, which contradicts $c \ne 0$. Consequently such $t$ doesn't exist and $f(x) = 0$ for all $x \in \mathbb{R}$.