Solve $f(x^2+y+f(y))=2y+f(x)^2$ over $\mathbb{R}$

Let me start with some simple observations building on what you have found so far. Since $f(x)=0$ implies $x=0$ and $f$ is surjective, we must have $f(0)=0$. Setting $y=0$ then gives $f(x^2)=f(x)^2$. Letting $x=\sqrt{a}$ we can rewrite the functional equation as $$f(a+y+f(y))=2y+f(a)$$ where $a$ is required to be nonnegative. We also get that $f(a)=f(x)^2>0$ for any $a>0$.

Now fix your favorite positive number $r$ and let $s=2r+f(r)$. Setting $a=y=r$, we see that $f(s)=s$. Setting $a=0$ and $y=s$, we then see $f(2s)=2s$. Letting $a$ run through multiples of $s$ and $y=s$, we conclude that $f(ns)=ns$ for all $n\in\mathbb{N}$.

Now suppose there exists $c\geq 0$ is such that $f(c)>c$. We have by induction that $f(m(c+f(c)))=2mc$ for all $m\in\mathbb{N}$. In particular, let us choose $m$ large enough so that $m(f(c)-c)>s$. Let us write $d=m(c+f(c))$, so we have $d-f(d)>s$. We can then find $n\in\mathbb{N}$ such that $$f(d)+d<ns<2d.$$ Setting $a=ns-f(d)-d$ and $y=d$, we get $$f(ns)=2d+f(a)>2d>ns.$$ But $f(ns)=ns$, so this is a contradiction.

Thus no such $c$ can exist, and we have $f(c)\leq c$ for all $c\geq 0$. On the other hand, if $f(c)<c$, then $f(c+f(c))=2c>c+f(c)$, which we know is impossible. Thus $f(c)=c$ for all $c\geq 0$.

Finally, suppose $d<0$ and $f(d)\neq d$. Since $f(d)^2=f(-d)^2=(-d)^2$, we must have $f(d)=-d$. But then the functional equation with $y=d$ gives $f(a)=2d+f(a)$, which is impossible since $d<0$.

So we conclude that $f(x)=x$ for all $x\in\mathbb{R}$. This function does indeed satisfy the functional equation, so it is the only solution.