By the [Euler–Maclaurin formula][1] (with $p=4$, $m=0$, and $g(x):=\frac1n\,f^2(\frac xn)$ in place of $f(x)$ there in the formula),
$$d_n(f):=\int_0^1 f^2(y) \, dy - \frac{1}{n}\sum_{i=1}^n f^2(i/n) \\
=-\frac1{2n}\,f^2(1)-\frac1{6n^3}\,f(1)f'(1)+O(n^{-4})\le O(n^{-4})$$
for each $f\in\mathcal F$. 

A more careful analysis may show that  
$$\sup_{f\in\mathcal F}d_n(f)\le O(n^{-4}).$$


  [1]: https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula#The_formula