The mathematics in your post is correct:
$$\text{from} \qquad F_e = -\frac{d U_r}{dr} \qquad \text{and} \qquad F_e = \frac{dW}{dr} \qquad \text{it follows} \qquad \boxed{dW = -dU_r} \tag 1$$
As you already mentioned, the work $W$ done by a conservative force equals the negative of the change in potential energy $\Delta U$
$$W = -\Delta U \tag 2$$
What is seems to me is that you are confused how come $dW$ in Eq. (1) and $W$ in Eq. (2) mean the same thing. Note that the well known equation for work
$$W = \int \vec{F} \cdot d\vec{r} \tag 3$$
actually comes from
$$dW = \vec{F} \cdot d\vec{r} \tag 4$$
You can read the Eq. (4) as follows: a force $\vec{F}$ over infinitesimal small displacement $d\vec{r}$ does infinitesimal small work $dW$.
Responding to your edit, you are getting the wrong sign because independent variable $r$ in
Work done by electric field to move a charge from r to $\infty$ :
$$W=-\Delta U_r=\int_{r}^{\infty}\frac{kQq}{r^2}dr$$
is lower bound of the integral. Remember that by definition, if function $f$ is continuous on interval $[a,x]$ then
$$F(x) = \int_{a}^{x} f(t) dt \quad \rightarrow \quad F'(x) = f(x)$$
But in your case you have
$$F(x) = \int_{x}^{a} f(t) dt = - \int_{a}^{x} f(t) dt \quad \rightarrow \quad -F'(x) = f(x)$$
The expression for work done by electric field to move a charge from $r$ to $\infty$ that you found
$$U_r = \frac{k Q q}{r}$$
$$W = U_r$$
is correct. But when you take derivate of that expression to find force $F_e$, you have to take into account that $r$ was actually lower bound.
Best Answer
Assuming the charge begins and ends at rest, or moves between two points at constant velocity so that there is no change in kinetic energy, then the increase in the magnitude of the EPE is equal to the work done on the charge by an external agent against the direction of the electric field. An example is a battery that does work (converts chemical energy to electrical potential energy) to separate charge at its terminals increasing the electrical potential and electrical potential energy of the charge.
Under the same assumption, a decrease in the magnitude of the EPE is equal to the work done by the electric field on the particle. An example is the work done by an electric field moving charge through a resistor in a circuit. There is a drop in voltage (drop in electrical potential) across the resistor.
Hope this helps.