Fiber bundles can have nontrivial topology, but for this question, we might as well work in a patch where it's trivial. Also, this answer uses classical fields to avoid possible technical issues when dealing with field operators (or distributions) on a Hilbert space.
For any analytic function $f(x)$, we have the identity
$$
f(x+c)=e^{c\cdot\partial}f(x)
\tag{1}
$$
where $\partial$ is the derivative with respect to $x$ and $c$ is independent of $x$. We might be tempted to write this as
$$
f(y)=e^{(y-x)\cdot\partial}f(x),
\tag{2}
$$
but that's not correct unless we interpret it like this:
$$
f(y)=\Big[e^{(y-x_0)\cdot\partial}f(x)\Big]_{x_0=x}.
\tag{3}
$$
Either way, it doesn't matter whether $f$ is a component of a gauge field (in flat or curved spacetime) or a component of a spinor field (in a gauge theory or not). In other words, it doesn't matter what the function $f$ represents. Equations (1) and (3) are identities that hold for any analytic function $f$. By the way, regarding the distinction between analytic and smooth, see https://en.wikipedia.org/wiki/Non-analytic_smooth_function.
In contrast, covariant derivatives are used to define new fields that preserve certain properties, which can facilitate the construction of gauge-invariant lagrangians. Consider a spinor field $\psi$ coupled to an abelian gauge field $A$. I'm using an abelian gauge field in this example so that we don't need to worry about path-ordering. Suppose we define a new field $\psi'$ by
$$
\psi'(x+c)=\psi(x+c)-\exp\left(i\int_x^{x+c} dy\cdot A(y)\right)\psi(x).
\tag{4}
$$
Under a gauge transform
\begin{gather}
\psi(x)\to e^{i\theta(x)}\psi(x)
\\
A_\mu(x)\to A_\mu(x)+\partial_\mu\theta(x),
\tag{5}
\end{gather}
the field $\psi'$ transforms as
$$
\psi'(x)\to e^{i\theta(x)}\psi'(x),
\tag{6}
$$
just like $\psi$ does. For infinitesimal $c$, equation (4) becomes
\begin{align}
\psi'(x+c)
&=\psi(x)+c\cdot\partial\psi(x)-ic\cdot A\psi(x) + O(c^2) \\
&=\psi(x)+c\cdot D\psi(x) + O(c^2)
\tag{7}
\end{align}
with $D_\mu=\partial_\mu-iA_\mu$. By the way, the finite version (4) is used in lattice gauge theory, where derivatives are replaced by finite differences.
I don't have a copy of Greiner's book on hand, but I'm guessing it's using the same symbol $\psi$ for both functions, writing $\psi$ instead of $\psi'$ on the left-hand side of (7). Yeah, that's careless, but it's also common: people often use the same symbol for different-but-related things. It's a compromise to avoid a proliferation of primes, tildes, hats, subscripts/superscripts, and other decorations when several different related quantities are involved. I don't know if that's what Greiner's book is doing here, but it's a plausible guess.
Best Answer
Note that the finite transformation of: $$ W^a_\mu \to W^a_\mu + \frac{1}{g} \partial_\mu \theta^a + \epsilon^{abc} \theta^b W^c_\mu $$ is: $$ W^a_\mu t^a \to g W_\mu^a t^a g^{-1} + \frac{i}{g} \partial_\mu g \tag{1} $$ where: $$ g = \exp(-i \theta^a t^a) \;\;\; \text{and} \;\;\; [t^a,t^b] = i \epsilon^{abc} t^c $$ Thus, the first term on the right-hand side of equation $(1)$ transforms under the adjoint representation of the Lie group. The second term does not transform under the adjoint representation, but it should be easy to verify that the transformed gauge field still takes values in the Lie algebra (hint: looking at infinitesimal transformations is the easiest method to verify this).
In case you want more information on the adjoint representation of the Lie group, it might be worth looking at this question.