Numerical experiments suggest that
$$A_2(n) := \sum_{k=1}^{n-1} k^2\sigma(k)\sigma(n-k) = \frac{n^2}{8}\sigma_3(n) - \frac{4n^3-n^2}{24}\sigma(n).$$
PS. In fact, it directly follows from the quoted Touchard and Ramanujan identities.

A couple of similar identities:
$$A_1(n):=\sum_{k=1}^{n-1} k\sigma(k)\sigma(n-k) = \frac{5n}{24}\sigma_3(n) - \frac{6n^2-n}{24}\sigma(n).$$
$$A_3(n):=\sum_{k=1}^{n-1} k^3\sigma(k)\sigma(n-k) = \frac{n^3}{12}\sigma_3(n) - \frac{3n^4-n^3}{24}\sigma(n).$$

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**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:
$$\mathcal{A}_n(x) = \frac{1}{1-x}\mathcal{A}_n(\frac{x}{x-1}).$$