Let the position vectors of $A, B, C,$ and $D$ be $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$, respectively.

Then the position vector of $E$ is:

$$ \vec{E} = \frac{\vec{a} + \vec{c}}{2} $$

And the position vector of $F$ is:

$$ \vec{F} = \frac{\vec{b} + \vec{d}}{2} $$

Now, we are given the equation:

$$ (\overrightarrow{AB} - \overrightarrow{BC}) + (\overrightarrow{AD} - \overrightarrow{DC}) = k\overrightarrow{FE} $$

We can rewrite this equation using the position vectors:

$$ (\vec{b} - \vec{a} - (\vec{c} - \vec{b})) + (\vec{d} - \vec{a} - (\vec{c} - \vec{d})) = k(\vec{E} - \vec{F}) $$

Simplifying the equation, we get:

$$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = \frac{k}{2}(2\vec{E} - 2\vec{F}) $$

Now substitute $\vec{E}$ and $\vec{F}$ expressions we found earlier:

$$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = \frac{k}{2}\left(2\left(\frac{\vec{a} + \vec{c}}{2}\right) - 2\left(\frac{\vec{b} + \vec{d}}{2}\right)\right) $$

Simplifying the equation:

$$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = -\frac{k}{2}(\vec{b} + \vec{d} - \vec{a} - \vec{c}) $$

Since both sides of the equation are equal:

$$ k = -4 $$