Consider the following to $q$-series \begin{align*} f(q):&=\sum_{k=1}^{\infty} \frac{(-1)^{k-1}(1 + q^k)\,q^{\binom{k + 1}2}}{(1 - q^k)^2} \qquad \text{and} \\ g(q):&=\frac1{\prod_{j=1}^{\infty}(1-q^j)^3}\cdot \sum_{k=1}^{\infty} (-1)^{k-1}(1^2+2^2+\cdots+k^2)\,q^{\binom{k + 1}2}. \end{align*}

I am interested in these two alternatives generating functions for sum of divisors.

QUESTION. Can you provide a direct combinatorial proof for $f(q)=g(q)$? I know other ways.

Remark. If $\sigma_1$ is the sum-of-divisors function, then we note a cute byproduct: $$\sum_{k=0}^n(-1)^k(2k+1)\cdot \sigma_1\left(\binom{n+1}2-\binom{k+1}2\right) =(-1)^{n-1}(1^2+2^2+\cdots+n^2).$$ It is also amusing the the sum of the arguments in the $\sigma_1$ function themselves satisfy $$\sum_{k=0}^n \left(\binom{n+1}2-\binom{k+1}2\right) =1^2+2^2+\cdots+n^2.$$