Let $P(x)$ be the asertion \[2f (f (x)) = (x^2 - x)f (x) + 4 - 2x\] If $f(2)=a$ , $P(2)$ gives $f(a)=a$ then $P(a)$ gives $(a-1)(a^2-4)=0$ so $f(2)$ can be $2,1$ or $-2.$ Set $f(1)=b$ then $P(1)$ gives $f(b)=1$ and $P(b)$ gives $b=1$ or $4.$

We need now also consider $f(0) = c$. Then $P(0)$ implies $f(c) = 2$, and $P(c)$ implies $2f(2) = 2(c-1)^2 + 2 \geq 2$. This eliminates the possibility $f(2) = -2$. If we had $f(2) = 1$, that will imply $c=1$, so $f(1) = 2$, absurd, since $f(1)$ can only be $1$ or $4$. Therefore $\boxed{f(2) = 2}$ (and then $c=0$ or $c=2$, but $c=0$ leads to $2 = f(0) = 0$, absurd, so $c=2$). An example with $f(1) = 4$ is $f(x) = \frac {4} {x}$ for $x\neq 0$, and $f(0)=2$. Let someone else find an example with $f(1) = 1$.

Since no-one volunteered the other example, here is one: $f(x) = \frac {4} {x}$ for $x\not \in \{0,1,4\}$, and $f(0)=2$, $f(1)=1$, $f(4) = -\frac {2} {3}$. The motivation behind is as follows. The formula $\varphi(x) = \frac {4} {x}$ is excellent, since it satisfies the functional equation for any $x\neq 0$, so let's try to keep as much of it as possible. Since we need $f(1) = 1$ (while the formula offers $\varphi(1) = 4$), it means also $f(4)$ needs have a different value than that of $\varphi(4) = 1$. Since $\varphi$ is an involution, maybe we can just get away with this little change. So denote $f(4)=d$; then $P(4)$ implies $2f(d) = 12d + 4 - 8$, i.e. $f(d) = 6d - 2$. Now, since $d=0$ and $d=1$ are both leading to an immediate contradiction, it means we need have $f(d) = \varphi(d) = \frac {4} {d}$. The ensuing equation is $\frac {4} {d} = 6d-2$, with roots $d=1$ and $d=-\frac {2} {3}$, therefore yielding the (only) convenient value $f(4) = -\frac {2} {3}$.