Gauge covariant derivative of a creation operator Suppose we define the (gauge) covariant derivative or as
$$\tilde{\nabla}=\nabla+ie\textbf{A},$$
where the vector potential $\textbf{A}$ has a matrix structure where only the diagonal has nonzero entries, but not all entries on the diagonal are the same. In calculating the covariant derivative of a product of an annihilation and a creation operator it appears that
$$\tilde{\nabla}(\psi\psi^\dagger)=(\nabla\psi+ie\textbf{A}\psi)\psi^\dagger+\psi(\nabla\psi^\dagger-ie\psi^\dagger\textbf{A}),$$
where there's a minus sign appearing in the second term. This suggests that the covariant derivative works differently when acting on a creation operator, namely:
$$\tilde{\nabla}\psi^\dagger=\psi^\dagger\tilde{\nabla}^\dagger=\nabla\psi^\dagger-ie\psi^\dagger\textbf{A}.$$
Question: Why does the covariant derivative act differently on the creation operator, why is it conjugated? If it's any help, the annihilation and creation operators are defined as
$$\psi=\begin{pmatrix}\psi_\uparrow\\\psi_\downarrow\\\psi_\uparrow^\dagger\\\psi_\downarrow^\dagger\end{pmatrix},\qquad\psi^\dagger=\begin{pmatrix}\psi_\uparrow^\dagger&\psi_\downarrow^\dagger&\psi_\uparrow&\psi_\downarrow\end{pmatrix}.$$
 A: Given that the regular (non-covariant) derivative of the adjoint satisfies
$$
\nabla \psi^\dagger = \left(\nabla \psi \right)^\dagger
$$
one likewise expects and defines
$$
\tilde{\nabla}\psi^\dagger = \left( \tilde{\nabla}\psi \right)^\dagger = \left( \nabla \psi +ie\bf{A}\psi \right)^\dagger = \nabla \psi^\dagger -ie\bf{A} \psi^\dagger 
$$
A: Let us look at this prolem from a (relativistic) field theory perspective.
The Hamiltonian for $\psi$ must contain a term of the form
$$ \nabla\psi^\dagger\cdot\nabla\psi$$
due to Lorentz invariance. Assuming $\psi$ to transform in a representation of U(1). Resulting in the following simultanious transformations:
$$\begin{aligned}\psi\rightarrow U(x)\psi \\\\\psi^\dagger \rightarrow \psi^\dagger U^\dagger(x)\end{aligned}$$
where U(x) is a unitary (1x1) matrix describing the result of gauge transforming $\psi$.
It is then easy to verify the gauge invariance of $\psi^\dagger \psi$, using $U^\dagger(x)U(x)=1$. In the case of U(1) we can write $U(x)=e^{i\xi(x)}$ and therefore $U^\dagger(x)=e^{-i\xi(x)}$, where U is the fundamental representation and $U^\dagger$ is in the anti-fundamental one.
The above term in the Hamiltonian is somewhat problematic however, since 
$$ \nabla [\psi U(x)] = \nabla \psi U(x)+\psi \nabla U(x)$$
contains the nasty inhomogeneos $\psi \nabla U(x)$ term.
To solve this problem, the covariant derivative was introduced as a variant of $\nabla$ that commutes with the transformation, that is under a gauge transformation
$$ \tilde{\nabla} \psi \rightarrow U(x)(\tilde {\nabla}\psi) $$
we can take this as a defining property of $\tilde{\nabla}$. Notice however that the exact form of U(x) is determined by the transformation of $\psi$ and thus depends on the representation of $\psi$. In order to learn more about $\tilde{\nabla}$ we might express it in terms of $\nabla$ and `some other object' that we will call X for now.
$$ \tilde{\nabla}=\nabla+X$$
we would then like it to satisfy
$$ \tilde{\nabla}[\psi U(x)]=\tilde{\nabla}[\psi] U(x)$$
expanding both in terms of the ordinary derivative:
$$ (\nabla+X)[\psi U(x)]=\nabla\psi U(x) + \psi \nabla U(x) + X\psi U(x)$$
and we would like this to be equivalent to 
$$ [\nabla\psi] U(x) + X\psi U(x)$$
without the nasty inhomogeneity. Notice how the offending term could be expressed as 
$$\nabla U(x)=i\nabla[\xi(x)] U(x)$$
Simmilarly:
$$\nabla U^\dagger(x)=-i[\nabla\xi(x)] U^\dagger(x)$$
It is this sign difference that confuses you. To illuminate, express X in terms of A. Since the physical part of A remains unchanged under a transformation 
$$ A \rightarrow A+\nabla\xi(x)$$
(when accompanied by a suitable transformation of V).
A suitable choice of X is
$$ X=\pm iA$$
It is however impossible to use the same expression for X when dealing with the anti-fundamental, to get rid of both the $\nabla U$ and $\nabla U^\dagger$ term using the same transformation. In stead, we need to pick a different X, depending on the representation it is acting on
$$\begin{aligned} XU(x)=\pm iAU(X)\\\\ XU^\dagger(x)=\mp iAU^\dagger(X)\\\\\end{aligned}$$
More often, X is expressed in terms of the relevant generator:
$$ X=A\tau $$
where $\tau$ is the generator of the group, in the prepresentation it is acting on. In the case of U(1), $\tau$ can be either equal to $i$ or equal to $-i$, depending on your conventions, either can be assignt to the fundamental representation, of the anti-fundamental one.
edit fixed some typography
further edit
when acting on $\psi\psi^\dagger$ in stead, simply use the Leibnitz rule (which still holds for the covariant derivative) and when expanding, let the two different X operators act from both sides, make sure the are expanded in different generators, again resulting in the sign difference.
tldr edit
TLDR: Your definition of the covariant derivative is not the actual definition, it is a consequence thereof. Indeed, the actual covariant derivative can be found to act on $\psi$ differently from $\psi^\dagger$.
