Given,

$ABCD$ be a quadrilateral. If $E$ and $F$ are the mid points of the diagonals $AC$ and $BD$ respectively and $\left(\overrightarrow{AB}- \overrightarrow{BC}\right)+\left(\overrightarrow{AD}-\overrightarrow{DC}\right)=k \overrightarrow{FE}$,

Now plotting the diagram we get,

Now let position vector of $A, B, C$ and $D$ are $\overrightarrow{a,} \overrightarrow{b,} \overrightarrow{c}$ and $\overrightarrow{d}$ respectively

Now by midpoint formula,

Position vector of $E=\frac{\overrightarrow{OC}+\overrightarrow{OA}}{2}=\frac{\overrightarrow{c}+\overrightarrow{a}}{2}$

And Position vector of $F=\frac{\overrightarrow{b}+\overrightarrow{d}}{2}$

Now, $\left(\overrightarrow{AB}-\overrightarrow{BC}\right)+\overrightarrow{AD}-\overrightarrow{DC}$

$\Rightarrow \overrightarrow{b}-\overrightarrow{a}-\left(\overrightarrow{c}-\overrightarrow{b}\right)+\overrightarrow{d}-\overrightarrow{a}-\left(\overrightarrow{c}-\overrightarrow{d}\right)$

$\Rightarrow 2\overrightarrow{b}-2\overrightarrow{a}-2\overrightarrow{c}+2\overrightarrow{d}$

$\Rightarrow 2\left(\overrightarrow{b}+\overrightarrow{d}\right)-2\left(\overrightarrow{a}+\overrightarrow{c}\right)$

$\Rightarrow 4\left[\frac{\overrightarrow{b}+\overrightarrow{d}}{2}-\frac{\overrightarrow{a}+\overrightarrow{c}}{2}\right]=4\left[\overrightarrow{OF}-\overrightarrow{OE}\right]$

$\Rightarrow 4\overrightarrow{EF}=-4\overrightarrow{FE}$

$\therefore k=-4$