Do derivatives anticommute with Grassmann variables and complex numbers in a many-body path integral?

Jane, $\partial_\tau$ is clearly a derivative with respect to a bosonic time $\tau$, so it commutes with everything else (except for functions of $\tau$ itself, with which it has a nonzero commutator), rather than anticommutes. Only if both objects have a fermionic character (if both of them are Grassmann-odd), they anticommute with one another (or they have an anticommutator that can be evaluated).

There is no sign error in the formulae, however. You asked a good question: how do you get from $$-\partial_\tau \bar\psi \psi$$ to $$+\bar\psi\partial_\tau \psi$$ One needs a bit of patience to answer this question. Note that in the two expressions, a different variable is differentiated. In the first one, it's $\bar\psi$ that is differentiated; in the second one, it's $\psi$.

You can't just move derivatives around. Even for bosonic functions, $uv'$ is something else than $u'v$, isn't it?

So the two expressions are not "obviously equal", not even up to a sign, and to convert one to the other, you must carefully integrate by parts. Note that $$\partial_\tau (\bar\psi \psi) = \partial_\tau\bar\psi \psi + \bar\psi\partial_\tau\psi. $$ This "Leibniz rule" proceeded just like for the derivative of products of bosonic factors because I had to bubble $\partial_\tau$ through the $\psi$'s, and $\partial_\tau$ is a bosonic object. If I were writing down a Leibniz rule for a Grassmannian derivative, I would have to change the sign everytime the derivative would bubble through a Grassmann-odd factor.

But here we deal with bosonic $\tau$-derivatives so the Leibniz rule is just like it has always been. So it implies that up to a total derivative - namely the left-hand side $\partial_\tau (\bar\psi \psi)$ that integrates to zero over the periodic Euclidean time - the two terms on the right hand side are opposite to one another. That's where the minus sign came from.