I think I am starting to understand what is going on on. For any Dirichlet character $\chi$, the Eisenstein series
$$\hat{E}_{k,\chi}(z)=-\frac{B_{k,\chi}}{2k}+\sum_{n=1}^{\infty}\left(\sum_{d|n}\chi(d)d^{k-1}\right)q^n$$
is an eigenform of the space $M_k(N,\chi)$. Thus, letting $\chi=\chi_2$ be the unique Dirichlet character modulo $2$ (i.e $\chi_2(n)=n\,\,\mathrm{mod}\,\,2$) we get that
\begin{align*} \hat{E}_{k+1,\chi_2}(z)&=-\frac{B_{k+1,\chi_2}}{2(k+1)}+\sum_{n=1}^{\infty}\left(\sum_{d|n}\chi_2(d)d^{k}\right)q^n\\ &=-\frac{B_{k+1,\chi_2}}{2(k+1)}+\sum_{n=1}^{\infty}\sigma^{(o)}_{k}(n)q^n \end{align*}
Thus, we can differentiate our functions $f_k(z)$ repeatedly $k$ times to get that
\begin{align*} f^{(k)}_k(z)&=\frac{d^k}{dz^{k}}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sigma_k^{(o)}(n)}{n^k}e^{z\pi i n}\\ &=(i\pi)^k\sum_{n=1}^{\infty}(-1)^{n-1}\sigma_k^{(o)}(n)e^{z\pi i n}\\ &=-(i\pi)^k\sum_{n=1}^{\infty}\sigma_k^{(o)}(n)e^{(z-1)\pi i n}\\ \end{align*}
and thus
$$\frac{-f^{(k)}_k(z+1)}{(i\pi)^k}=\hat{E}_{k+1,\chi_2}(z)+\frac{B_{k+1,\chi_2}}{2(k+1)}$$
In a way this explains the connection we are seeing, but in another it makes it even more mysterious since there is no reason that the iterated integral of an eigenform should retain its nice properties.