Numerical experiments suggest that $$\sum_{k=1}^{n-1} k^2\sigma(k)\sigma(n-k) = \frac{n^2}{8}\sigma_3(n) - \frac{4n^3-n^2}{24}\sigma(n).$$$$A_2(n) := \sum_{k=1}^{n-1} k^2\sigma(k)\sigma(n-k) = \frac{n^2}{8}\sigma_3(n) - \frac{4n^3-n^2}{24}\sigma(n).$$ PS. In fact, it directly follows from the quoted Touchard and Ramanujan identities.
A couple of similar identities: $$\sum_{k=1}^{n-1} k\sigma(k)\sigma(n-k) = \frac{5n}{24}\sigma_3(n) - \frac{6n^2-n}{24}\sigma(n).$$$$A_1(n):=\sum_{k=1}^{n-1} k\sigma(k)\sigma(n-k) = \frac{5n}{24}\sigma_3(n) - \frac{6n^2-n}{24}\sigma(n).$$ $$\sum_{k=1}^{n-1} k^3\sigma(k)\sigma(n-k) = \frac{n^3}{12}\sigma_3(n) - \frac{3n^4-n^3}{24}\sigma(n).$$$$A_3(n):=\sum_{k=1}^{n-1} k^3\sigma(k)\sigma(n-k) = \frac{n^3}{12}\sigma_3(n) - \frac{3n^4-n^3}{24}\sigma(n).$$
ADDED. A recurrent formula for $A_d(n)$ with an odd $d$ can be obtained from the observation: \begin{split} A_d(n) & := \sum_{k=1}^{n-1} k^d\sigma(k)\sigma(n-k) \\ &= \sum_{k=1}^{n-1} (n-k)^d\sigma(k)\sigma(n-k) \\ &= \sum_{i=0}^d \binom{d}{i} n^{d-i} (-1)^i A_i(n). \end{split} implying that \begin{split} A_d(n) &= \frac{1}{2} \sum_{i=0}^{d-1} \binom{d}{i} n^{d-i} (-1)^i A_i(n) \\ &=\frac{1}{d+1} \sum_{i=0}^{d-1} \binom{d+1}{i} n^{d-i} (-1)^i A_i(n). \end{split} However, to use this formula one would need to compute $A_t(n)$ for even $t<d$ by other means.
It also follows that the generating function: $$\mathcal{A}_n(x) := \sum_{d=0}^{\infty} \frac{A_d(n)}{n^d}x^d$$ satisfied the functional equation: $$\mathcal{A}_n(x) = \frac{1}{1-x}\mathcal{A}_n(\frac{x}{x-1}).$$