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Find all functions $f$ from the reals to the reals such that

\[f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)\]

for all real $x,y$.

Plug in $y=-f(x)$ to get that $f(0)=2x+f(\text{something})$, so $f$ is surjective. So, there exists $t$ so that $f(t)=0$.

Plug $x=t$. We get that $f(y)=2t+f(f(y)-t)$. Set $f(y)=x$ here, and due to surjectivity $x$ ranges over all reals. So, $x=2t+f(x-t)$, equivalently $x+t=2t+f(x)$. So, general solution is $f(x)=x+a$ for some real constant $a$, and plugging in we see the functional equation works.