Solution:

From the given equation (1) we obtain, for $y=-2x$: $g(f(-x))=f(x)$ (2). Thus (1) becomes $f(-x-y)=f(x)+(2x+y)g(y)$ (3), and with $x=0$ this becomes $f(-y)=f(0)+y g(y)$ (4).

Replacing $f$ in (3) according to (4) yields, after subtracting $f(0)$:
$$(x+y)g(x+y) = (2x+y)g(y) - xg(-x)$$
(5).

With $y=0$ this gives, for $x \neq 0$: $g(x)+g(-x)=2g(0)$ (6), which however also holds for $x=0$, hence for all $x \in \mathbb{R}$.

We set $h(x):=g(x)-g(0)$. Then $h(0)=0$ holds, as well as, by (6): $h(-x)=-h(x)$ (6*), and, using (5) and (6*):
$$(x+y)h(x+y) = (2x+y)h(y) + x h(x)$$
(7).

Interchanging $x$ and $y$ in (7) and equating the results yields $x h(y) = y h(x)$ (8). For $x, y \neq 0$ it follows that $h(x)/x = h(y)/y = \text{const}$. Hence $h(x) = c \cdot x$, which obviously also holds for $x=0$. With $g(0)=b$ we obtain $g(x)=c x + b$ for all $x \in \mathbb{R}$.

With $f(0)=a$ it follows from (4) that $f(x) = a - b x + c x^2$. (2) yields $c(c x^2 + b x + a) + b = c x^2 - b x + a$, and comparing coefficients we must have $c^2 = c$, $b c = -b$ as well as $a c + b = a$.

Case 1: $c=0$. This gives $b=a=0$ and $f(x)=0$, $g(x)=0$, so that (1) is satisfied.

Case 2: $c=1$. Then $b=-b$, so $b=0$, and $a$ is arbitrary. This gives $f(x)=x^2 + a$, $g(x)=x$. For this pair of functions too, substituting into (1) yields a true statement. Hence the two solution pairs $(f, g) = (0, 0)$ and $(f, g) = (x^2 + a, x)$ with $a \in \mathbb{R}$ exist.