Solution:
For all positive integers $n$, let $\omega(n) = f(n) - f(n - 1)$. We claim that $\omega(n)$ is the largest odd divisor of $n$ for all $n > 0$. Indeed, for all positive integers $k$, we have
$$\omega(2k) = f(2k) - f(2k - 1) = f(k) + k^2 - (f(k - 1) + k^2) = f(k) - f(k - 1) = \omega(k)$$
and
$$\omega(2k + 1) = f(2k + 1) - f(2k) = f(k) + (k + 1)^2 - (f(k) + k^2) = 2k + 1.$$
Inducting on the positive integers implies that $\omega(n)$ is indeed the largest odd divisor of $n$.

We can now rewrite the sum as
$$\sum_{n = 1}^{31} \frac{f(n)}{n(n + 1)} = \sum_{n = 1}^{31} \left( \frac{f(n)}{n} - \frac{f(n)}{n + 1} \right ) = \left( \sum_{n = 1}^{30} \frac{f(n) - f(n - 1)}{n} \right ) - \frac{f(31)}{32}.$$
Note that by using the original recursive definition, we can compute
$$f(31) = 16^2 + 8^2 + 4^2 + 2^2 + 1^2 = 341.$$
Moreover, we also see that $\frac{f(n) - f(n - 1)}{n} = \frac{\omega(n)}{n} = 2^{-\nu_2(n)}$, where $2^{\nu_2(n)}$ is the largest power of $2$ dividing $n$. Thus, our desired sum is
$$\left( \sum_{n = 1}^{31} 2^{-\nu_2(n)} \right ) - \frac{341}{32} = 16 \cdot 2^{-0} + 8 \cdot 2^{-1} + 4 \cdot 2^{-2} + 2 \cdot 2^{-3} + 1 \cdot 2^{-4} - \frac{341}{32} = \frac{341}{32}.$$

Solution 2:
From the original recursion, for all positive integers $n$, we have
$$\frac{f(2n)}{2n(2n + 1)} = \frac{f(n) + n^2}{2n(2n + 1)} = \frac{f(n)}{2n(2n + 1)} + \frac{n}{2(2n + 1)}$$
and
$$\frac{f(2n + 1)}{(2n + 1)(2n + 2)} = \frac{f(n) + (n + 1)^2}{(2n + 1)(2n + 2)} = \frac{f(n)}{(2n + 1)(2n + 2)} + \frac{n + 1}{2(2n + 1)}.$$
Adding these two equations gives
$$\frac{f(2n)}{2n(2n + 1)} + \frac{f(2n + 1)}{(2n + 1)(2n + 2)} = \frac{f(n)}{2n(n + 1)} + \frac{1}{2}.$$
Thus, if $T(n) = \sum_{k = 1}^n \frac{f(k)}{k(k + 1)}$, we have
$$T(2n + 1) = \sum_{k = 1}^{2n + 1} \frac{f(k)}{k(k + 1)} = \frac{f(1)}{1 \cdot 2} + \sum_{m = 1}^n \left( \frac{f(2m)}{2m(2m + 1)} + \frac{f(2m + 1)}{(2m + 1)(2m + 2)} \right )$$
$$= \frac{1}{2} + \sum_{m = 1}^n \left( \frac{f(m)}{2m(m + 1)} + \frac{1}{2} \right )$$
$$= \frac{n + 1}{2} + \frac{1}{2} T(n).$$
Starting from $T(1) = 1$, we can compute $T(3) = \frac{5}{4}$, $T(7) = \frac{21}{8}$, $T(15) = \frac{85}{16}$, and finally, $T(31) = \frac{341}{32}$.