I think I am starting to understand what is going on on. For any Dirichlet character $\chi$, theThe generalized Eisenstein series
$$\hat{E}_{k,\chi}(z)=-\frac{B_{k,\chi}}{2k}+\sum_{n=1}^{\infty}\left(\sum_{d|n}\chi(d)d^{k-1}\right)q^n$$\begin{equation} \hat{E}_{k,\chi,\psi}(z):=\sum_{n=1}^{\infty}\left(\sum_{d|n}\psi(d)\chi(n/d)d^{k-1}\right)q^n\tag{1} \end{equation}
is anIs a eigenform ofover the space $M_k(N,\chi)$$M_k(RL,\chi\psi)$ for characters $\psi$ and $\chi$ with conductors $L$ and $R$ respectively, whenever $L>1$ and $\psi(-1)\chi(-1)=(-1)^k$. ThusMore generally, lettingfor any $\chi=\chi_2$$t>0$ the function $\hat{E}_{k,\chi,\psi}\left(z^t\right)$ is a modular form over $M_k(RLt,\chi\psi)$ according to this online textbook, Theorem 5.8.
We now set $\psi=1$ to be the trivial character and $\chi_=\chi_2$ to be the unique Dirichlet character modulo $2$. This means that $L=2>1$ and $\psi(-1)\chi(-1)=1$ so the conditions of (i.e1) are satisfied if $\chi_2(n)=n\,\,\mathrm{mod}\,\,2$$k$ is even, making $\hat{E}_{k,\chi_2,1}(z)$ a modular (eigen) form. Now, if $k'$ is any odd integer we get thatcan let $k=1-k'$ be an even integer, so
\begin{align*} \hat{E}_{k+1,\chi_2}(z)&=-\frac{B_{k+1,\chi_2}}{2(k+1)}+\sum_{n=1}^{\infty}\left(\sum_{d|n}\chi_2(d)d^{k}\right)q^n\\ &=-\frac{B_{k+1,\chi_2}}{2(k+1)}+\sum_{n=1}^{\infty}\sigma^{(o)}_{k}(n)q^n \end{align*}\begin{equation} \hat{E}_{1-k',\chi_2,1}(z)=\sum_{n=1}^{\infty}\left(\sum_{d|n}\chi_2(n/d)d^{-k'}\right)q^n \end{equation}
Thus, we can differentiate our functionsis a modular form of degree $f_k(z)$ repeatedly$1-k'$ over $k$ times to get$\Gamma_1(2)$. Since $\sum_{d|n}\chi_2(n/d)d^{-k'}=\frac{\sigma^{(o)}_{k}(n)}{n^k}$, this means that
\begin{align*} f^{(k)}_k(z)&=\frac{d^k}{dz^{k}}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sigma_k^{(o)}(n)}{n^k}e^{z\pi i n}\\ &=(i\pi)^k\sum_{n=1}^{\infty}(-1)^{n-1}\sigma_k^{(o)}(n)e^{z\pi i n}\\ &=-(i\pi)^k\sum_{n=1}^{\infty}\sigma_k^{(o)}(n)e^{(z-1)\pi i n}\\ \end{align*}\begin{equation} \hat{E}_{1-k,\chi_2,1}(z)=\sum_{n=1}^{\infty}\frac{\sigma^{(o)}_{k}(n)}{n^k}q^n\tag{2} \end{equation}
and thusRelating this to $f_{k}(z)$ is now a straightforward task. Namely, we see that
$$\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)}$$\begin{align*} f_k(2z+1)&=2i\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sigma_k^{(o)}(n)}{n^k}\exp\left(\frac{2\pi in(2z+1)}{2}\right)\\ &=-2i\sum_{n=1}^{\infty}\frac{\sigma_k^{(o)}(n)}{n^k}\exp\left(2\pi inz\right)\\ &=-2i\hat{E}_{1-k,\chi_2,1}(z) \end{align*}
In a way this explains the connection we are seeingThus, but in another it makes it even more mysterious since there is no reason that the iterated integral of antransformation $f_k(2z+1)i/2$ turns $f_k(z)$ into a eigenform should retain its nice properties.
Another (possibly more enlightening) way to think about what is going on is Working out that $f_k(z)$$f_k(2z+1)i/2$ is an eigenform just from the function generated byformulae given in the $q$-series $\sum_{d|n}\chi_2\left(\frac{n}{d}\right)d^{-k}$problem statement seems hard, and so perhaps for any choicebut also I am not aware of Dirichlet character $\chi$any general formula for Eisenstein series which yields the $q$-series generated by $\sum_{d|n}\chi\left(\frac{n}{d}\right)d^{-k}$ will have modular form-like properties,formulae given in the sense that
$$z^{\frac{1-k}{2}}g(z)+z^{\frac{k-1}{2}}g\left(\frac{1}{z}\right)=\mathrm{poly}(z)$$problem statement.
where $\mathrm{poly}(z)$ is a polynomial of degree at most $\frac{k+1}{2}$I find this stuff absolutely fascinating, and I'll be looking more into this and undoubtedly coming up with lots more fun stuff, so if anyone actually ends up reading this and cares about I'm writting then I will add more answers/edit this answer as time goes on to make the solution more complete. Thank you for your time.