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.