Answer: $f(x) = a(x + 2a)$, $g(x) = a(1 - 2a)(x + 2a)$, $a \in \mathbb{R}$.

(Solution by A. Goloubitskaya, L. Manzhulina.) First, we claim that $f$ takes the value $0$. If $f(0) = 0$, then there is nothing to prove. If $f(0) = b \neq 0$, then setting $x = 0$ in the given equality
$$f(x - 2f(y)) = xf(y) - yf(x) + g(x), \quad (1)$$
we obtain $f(-2f(y)) = -by + g(0)$. The right-hand side attains all real values since $b \neq 0$. So $f$ attains all real values including $0$.
Let $c$ be such that $f(c) = 0$. Set $y = c$ in (1): $f(x) = -cf(x) + g(x)$, whence $g(x) = (c+1)f(x)$. Therefore, (1) becomes
$$f(x - 2f(y)) = xf(y) + (c + 1 - y)f(x). \quad (2)$$
Let $a = f(c + 1)$. Substituting $c + 1$ for $y$ in (2), we obtain $f(x - 2a) = ax$, whence $f(x) = ax + 2a^2$. Since $f(c) = 0$, we have $ac + 2a^2 = 0$ and
$$g(x) = (c+1)(ax + 2a^2) = (-2a^2 + a)x - 4a^3 + 2a^2.$$
Easy verification shows that these functions $f$ and $g$ satisfy the given equality for all real $a$.