For any real number $z$, we set $a=f(z), b=z+f(0), n=f\left(\frac{a+b}{2}\right)$, and $m=f(0)-\frac{a+b}{2}$. Since $\mathbb{D}_{f}=\mathbb{R}$, $a, b, n$, and $m$ are well-defined. Substituting $x=0$, $y=z$ into $(*)$ yields $f(z+f(0))=f(f(z))$, so $f(a)=f(b)$ (I).
Substituting $x=m$ and $y=a$ or $y=b$ into $(*)$ yields $\left\{\begin{array}{l}f(f(m)+a)=2 m+f(f(a)-m) \\ f(f(m)+b)=2 m+f(f(b)-m)\end{array}\right.$, which, due to (I), can be summarized as $f(f(m)+a)=f(f(m)+b)$ (II).
Substituting $x=a$ or $x=b$ and $y=n$ into $(*)$ yields $\left\{\begin{array}{l}f(f(a)+n)=2 a+f(f(n)-a) \\ f(f(b)+n)=2 b+f(f(n)-b)\end{array}\right.$, which, due to (I), can be summarized as $2 a+f(f(n)-a)=2 b+f(f(n)-b)$ (III). Substituting $x=\frac{a+b}{2}, y=0$ into $(*)$ yields $f\left(f\left(\frac{a+b}{2}\right)\right)=a+b+f\left(f(0)-\frac{a+b}{2}\right)$, so $f(n)=a+b+f(m)$ or, in other words, $f(n)-a=f(m)+b$ and $f(n)-b=f(m)+a$. This, combined with (III), immediately gives $2 a+f(f(m)+b)=2 b+f(f(m)+a)$, from which, with (II), we conclude $a=b$, so $f(z)=z+f(0)$. This shows that any function $f$ that satisfies $(*)$ must have the form $f(x)=x+c \quad(c=$ const. ). Substituting into $(*)$ confirms that any $f$ of this form indeed satisfies the given equation: $f(f(x)+y)=x+y+2 c=2 x+f(f(y)-x)$.
Note: Additional properties of $f$, such as injectivity, monotonicity, or differentiability, should not be assumed but must be proven.