By "transform like a function of the coordinates," I mean that under an infinitesimal translation $x^\mu \to x^\mu + \epsilon^\mu$, to first order in $\epsilon^\mu$ the function $f(t,\mathbf x)$ becomes
$f(t,\mathbf x) \to f(t,\mathbf x) + \epsilon^\nu (\partial_\nu f)(t,\mathbf x)$
Suppose we start in the temporal gauge with $A_0 = 0$ and then perform a gauge transformation so that $A_0 = \partial_0 \lambda$ for some scalar field $\lambda(t,\mathbf x)$. If $\lambda$ transforms as a function of the coordinates under translations, then this becomes
$\partial_0 \lambda \to \partial_0 (\lambda + \epsilon^\mu \partial_\mu \lambda) = \partial_0 \lambda + \partial_0(\epsilon^\mu \partial_\mu \lambda)$
If $A_0 \neq 0$ before the gauge transformation and also transforms like an ordinary function, then it's not consistent
$A_0 + \partial_0 \lambda \to (A_0 + \partial_0 \lambda) + \epsilon^\nu \partial_\nu (A_0 + \partial_0 \lambda) + (\partial_0 \epsilon^\nu) (\partial_\nu \lambda)$
If $A_\mu$ instead transforms under translation as
$A_\mu \to A_\mu + \partial_\mu(\epsilon^\nu A_\nu)$
then consistency is restored:
$(A_\mu+\partial_\mu\lambda) \to (A_\mu+\partial_\mu\lambda) + \partial_\mu(\epsilon^\nu (A_\nu+\partial_\nu\lambda))$
If so, this would explain why the canonical momentum to position for the Dirac field includes the vector potential, whereas otherwise I can't see how to get the time derivative of the displacement to multiply $A_k$ in the transformed Lagrangian. Then when $x^\mu \to x^\mu + \epsilon^\mu$ we get
$\mathcal L = \bar\psi \gamma^\mu (i \partial_\mu - e A_\mu) \psi$
$\mathcal L \to (\bar\psi + \epsilon^\nu \partial_\nu \bar\psi) \gamma^\mu (i \partial_\mu - e A_\mu - e \partial_\mu(\epsilon^\nu A_\nu)) (\psi + \epsilon^\nu \partial_\nu \psi)$
$\mathcal L \to \dot \epsilon^k \bar\psi \gamma^0 (i \partial_k - e A_k) \psi + \cdots$
$p_k = \frac{\mathrm d L'}{\mathrm d \dot\epsilon^k} = \int \mathrm d^3\mathbf x \, \psi^\dagger (i \partial_k - e A_k) \psi $
