f(f(x) + y) = 2x + f(f(y) − x)

Step 1: surjectivity. Put $y=-f(x)$: $f(0)=2x+f\bigl(f(-f(x))-x\bigr)$, so
$$f\bigl(f(-f(x))-x\bigr)=f(0)-2x.$$
As $x$ ranges over $\mathbb{R}$, the right side takes every real value, so $f$ is surjective.

Step 2: a zero. Let $a$ be such that $f(a)=0$ and put $x=a$:
$$f(y)=2a+f\bigl(f(y)-a\bigr)\qquad\text{for all }y.$$

Step 3: linear form. Let $u=f(y)-a$. Since $f$ is surjective, $u$ ranges over $\mathbb{R}$. Then $f(y)=u+a$ and the relation becomes $u+a=2a+f(u)$, i.e.
$$f(u)=u-a\qquad\text{for all }u.$$
So $f(x)=x+c$ with $c=-a$.

Step 4: verification. For $f(x)=x+c$: the left side is $f(x+c+y)=x+y+2c$, and the right side is $2x+f(y+c-x)=2x+y+c-x+c=x+y+2c$ ✓.

Answer. $f(x)=x+c$ for any constant $c\in\mathbb{R}$.

Lesson. Surjectivity (via a substitution that makes one side independent of the argument) lets you replace $f(y)$ by a free variable: this is the standard route from a messy equation to a linear one.