A functional equation: $f(x, y + z) = f(x, y) + f(x + y, z)$

Take $y=0$ to get $f(x,z) = f(x,0) + f(x,z)$, so $f(x,0) = 0$.

$f(x,y) = g(x+y) - g(x)$ is a solution for arbitrary functions $g$. I don't know if those are all the solutions: they seem to be all the ones that are low-degree polynomials.

EDIT: Note that the transformation $f(x,y) = F(x,x+y)$ and change of variables $x = X, y = Y-X, z = Z - Y$ makes the equation $$ F(X,Z) = F(X,Y)+F(Y,Z)$$ Thus (taking, say, $Y=0$, $G(X) = F(X,0)$ and $H(Z) = F(0,Z)$) we have $F(X,Z) = G(X) + H(Z)$. But then $G(X) + H(Z) = G(X) + H(Y) + G(Y) + H(Z)$ which says $G(Y) = -H(Y)$. So we have $F(X,Z) = H(Z) - H(X)$, which transforms back to $f(x,y) = H(x+y) - H(x)$.

Following the beautiful idea of Robert Israel, we will show that the solutions of the equation are precisely functions of the form $f(x,y)=g(x+y)-g(x)$, where $g$ is an arbitrary function.

First, we plug $z = -y$ into the equation. This yields $$f(x, 0) = f(x, y) + f(x + y, -y).\tag{1}$$

In the case $y=0$, this tells us that $f(x,0)=0$ for all $x$, as Robert Israel already noticed. Using this fact in $(1)$, we have that $$f(x,y)=-f(x+y,-y)\tag{2}$$ must hold for all $x,y$ in order for $f$ to be a solution. We will now use this fact in the original equation. The original equation says that $$f(x, y) = f(x, y + z) - f(x + y, z)$$ holds for all $x,y,z$. Using $(2)$ twice we may rewrite this as $$f(x, y) = - f(x+y+z, - y - z) + f(x+y+z, -z)$$ Now, plug in $z=-x-y$ and get $$f(x, y) = - f(0, x) + f(0, x+y).$$ This means that if $f$ solves the original equation, we may define the function $g$ by $g(w)=f(0,w)$ and then $f(x,y)=g(x+y)-g(x)$ will hold. This shows that indeed the solutions of the equation are precisely of the form suggested by Robert Israel.