We start by calculating the dot product of $\overrightarrow{b}$ and $\overrightarrow{c}$:

$\overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{b} \cdot [(1-t)\overrightarrow{a} + t\overrightarrow{b}]$

Distributive law for dot products gives:

$=(1-t)(\overrightarrow{b} \cdot \overrightarrow{a}) + t(\overrightarrow{b} \cdot \overrightarrow{b})$

Since $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors and their angle is $60^{\circ}$, their dot product is $\frac{1}{2}$, and the dot product of a vector with itself is 1. So we have:

$=(1-t)\cdot\frac{1}{2} + t\cdot1$

$=\frac{1-t}{2} + t$

We are given that $\overrightarrow{b} \cdot \overrightarrow{c} = -\frac{1}{2}$, so we can set up the following equation:

$\frac{1-t}{2} + t = -\frac{1}{2}$

Solving for $t$, we get:

$t = -2$

Therefore, the answer is $\boxed{D: -2}$.