In Kohnen & Skoruppa's 1989 inventiones [paper][1], page 549, the operator $V_N: J_{k,1}^\text{cusp} \longrightarrow J_{k,N}^\text{cusp}$ is defined by the action
$$ \sum_{\substack{D<0,r \in \mathbb{Z}\\D\equiv r^2(4)}} c(D,r)\, e\left( \frac{r^2-D}{4}\tau+rz\right) \longmapsto \sum_{\substack{D<0,r \in \mathbb{Z}\\D\equiv r^2(4N)}} \left(\sum_{\substack{d|(r,N)\\D\equiv r^2(4Nd)}} d^{k-1} c\left(\frac{D}{d^2},\frac{r}{d}\right) \right) \,e\left( \frac{r^2-D}{4N}\tau+rz\right)  $$

Its adjoint (wrt Petersson inner product) $V_N^*: J_{k,N}^\text{cusp} \longrightarrow J_{k,1}^\text{cusp}$ is then given by 
$$ \sum_{\substack{D<0,r \in \mathbb{Z}\\D\equiv r^2(4N)}} c(D,r)\, e\left( \frac{r^2-D}{4N}\tau+rz\right) \longmapsto \sum_{\substack{D<0,r \in \mathbb{Z}\\D\equiv r^2(4)}} \left(\sum_{d|N} d^{k-2} \sum_{\substack{s(2d)\\s^2\equiv D(4d)}} c\left(\frac{N^2}{d^2}D,\frac{N}{d}s\right) \right) \,e\left( \frac{r^2-D}{4}\tau+rz\right)  $$

Then, in the Propostion (ii) [page 549-550] the action of $V_N^*V_N$ is given by 
$$ V_N^*V_N = \sum_{t|N}\Psi(t)t^{k-2}T\left( \frac{N}{t}\right), $$
where $\Psi(t)=t\prod_{t|p}\left(1+\frac{1}{p}\right)$ and $T(n)$ denotes the Hecke operator on $J_{k,1}^\text{cusp}.$

I am not able to follow how $V_N^*V_N$ is derived. I am not sure about action of $T(n)$ also. I first applied $V_N$ and then taking its coefficients to be $c(D',d')$ applied $V_N^*$ on it, but cannot proceed further. Any help? 


  [1]: https://doi.org/10.1007/BF01393889