Solution:
The square's corners must be at $(x, y), (x+1, y), (x+1, y+1)$, and $(x, y+1)$ for some $x$ and $y$. So,
$$\begin{aligned} f(x_{1}, y_{1}) & + f(x_{2}, y_{2}) + f(x_{3}, y_{3}) + f(x_{4}, y_{4}) \\ &= 2(x^{2} + 2x) + 2((x+1)^{2} + 2(x+1)) + 2(y^{2} + 4y) + 2((y+1)^{2} + 4(y+1)) \\ &= 4x^{2} + 12x + 6 + 4y^{2} + 20y + 10 \\ &= (2x+3)^{2} - 3 + (2y+5)^{2} - 15 \\ &\geq -18 \end{aligned}$$
This attains its minimum value of $-18$ when $x = -\frac{3}{2}$ and $y = -\frac{5}{2}$.