Solution:
First observe that $g(x)=0$ and $g(x)=x^{2}$ are solutions. Indeed, both are even functions. For $g(x)=0$, one can take $f(x)=0$ (or any function $f$ such that $f(x+f(x))=0$). For $g(x)=x^{2}$, one can take $f(x)=x$, because $(x+y)^{2}=x^{2}+y^{2}+2 x y$.

We now prove that these are the only solutions. Let $g$ be a solution of the problem and $f$ be an associated function. We start with a bunch of substitutions. Let $y=0$ which gives
$$g(f(x))=g(x)+g(0)$$
Let further $y=-f(x)$ in the original equation and use the parity of $g$ to get $g(0)=g(x)+g(f(x))-f(x) f(x+f(x))$. If we plug (1) in, we get
$$2 g(x)=f(x) f(x+f(x))$$
We replace $y$ by $f(y)$ in the original equation and use (1) to obtain $g(f(x)+f(y))=g(x)+g(y)+g(0)+f(y) f(x+f(x))$. The symmetry between $x$ and $y$ implies
$$f(y) f(x+f(x))=f(x) f(y+f(y))$$
If $f \equiv 0$, then $g \equiv 0$. So, if we assume that $g$ is not the constant $0$ function (which we know is a solution), then there exists $a$ such that $f(a) \neq 0$. Let $c=f(a+f(a)) / f(a)$. We have, with $y=a$ in (3), $f(x+f(x))=c f(x)$ and using (2) we obtain
$$2 g(x)=c f(x)^{2}$$
Since we assumed $g$ is not identically $0$, it holds $c \neq 0$. We plug (4) in the original equation and use $f(x+f(x))=c f(x)$ to obtain, after simplifying by $c$,
$$f(f(x)+y)^{2}=f(x)^{2}+f(y)^{2}+2 y f(x)$$
If one lets $x=y$ in (5), then one obtains
$$f(x)\left(f(x)\left(c^{2}-2\right)-2 x\right)=0$$
If $c^{2}=2$, then $f(x)=0$ for all $x \neq 0$. In particular, one must have $a=0$ and $f(0)=f(a) \neq 0$. But then $c=f(f(0)) / f(0)=0$ since $f(0) \neq 0$. Contradiction. Hence $c^{2} \neq 2$, and for every $x$,
$$f(x)=0 \quad \text{or} \quad f(x)=2 x /\left(c^{2}-2\right)$$
In particular, $f(0)=0$. Let $b:=2 /\left(c^{2}-2\right) \neq 0$ such that $f(x)=b x$ if $f(x) \neq 0$. Since $f(a) \neq 0$, we have $f(a)=a b$ and $a \neq 0$. Assume there is $t \neq 0$ such that $f(t)=0$, then, with $x=a$ and $y=t$ in (5), we get
$$f(a b+t)^{2}=a^{2} b^{2}+2 t a b$$
Because $t \neq 0$, we must have $f(a b+t)=0$ and thus $a b+2 t=0$. So there is at most one $t \neq 0$, such that $f(t)=0$, namely $t=-a b / 2$. But if such a $t$ exists, then $f(a b+t)=0$ and $a b+t \neq t$ and $a b+t \neq 0$, contradiction. So, $f(x)=b x$ for all $x$ and $g(x)=c b^{2} / 2 \cdot x^{2}$ for all $x$ by (4). We plug this in the original equation and deduce that $b=1$ and $c=2$ (using that $c \neq 0$). Therefore, we proved that if $g$ is not identically $0$, then $g(x)=x^{2}$ for every $x$.