jmc

Setting $y=0,$ we get $f(0)=0.$

Then setting $x=0,$ we get $$f(0)=2y[f(y)+f(-y)].$$Assuming $y\ne 0,$ we get $f(-y)+f(y)=0.$ Hence, $f(-y)=-f(y)$ for all $y.$

We can reverse the roles of $x$ and $y$ to get $$f(4xy)=2x[f(x+y)+f(y-x)],$$so $$2y[f(x+y)+f(x-y)]=2x[f(x+y)+f(y-x)].$$Hence, $$yf(x-y)-xf(y-x)=(x-y)f(x+y).$$Since $f(y-x)=-f(x-y),$ $$(x+y)f(x-y)=(x-y)f(x+y).$$We want to take $x$ and $y$ so that $x+y=5$ and $x-y=2015.$ Solving, we find $x=1010$ and $y=-1005.$ Then $$5f(2015)=2015f(5),$$so $f(2015)=\frac{2015f(5)}{5}=\boxed{1209}.$