The generalized Eisenstein series
\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 a eigenform over the space $M_k(RL,\chi\psi)$ for characters $\psi$ and $\chi$ with conductors $L$ and $R$ respectively, whenever $L>1$ and $\psi(-1)\chi(-1)=(-1)^k$. More generally, for any $t>0$ 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 character modulo $2$. This means that $L=2>1$ and $\psi(-1)\chi(-1)=1$ so the conditions of (1) are satisfied if $k$ is even, making $\hat{E}_{k,\chi_2,1}(z)$ a modular (eigen) form. Now, if $k'$ is any odd integer we can let $k=1-k'$ be an even integer, so
\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}
is a modular form of degree $1-k'$ over $\Gamma_1(2)$. Since $\sum_{d|n}\chi_2(n/d)d^{-k'}=\frac{\sigma^{(o)}_{k}(n)}{n^k}$, this means that
\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}
Relating this to $f_{k}(z)$ is now a straightforward task. Namely, we see that
\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*}
Thus, the transformation $f_k(2z+1)i/2$ turns $f_k(z)$ into a eigenform. Working out that $f_k(2z+1)i/2$ is an eigenform just from the formulae given in the problem statement seems hard, but also I am not aware of any general formula for Eisenstein series which yields the formulae given in the problem statement.
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.