We denote by $(a,b,c)$ the binary integral quadratic form $q(x,y)=ax^2+bxy+cy^2$, by $\text{Cl}(D)$ the set of $\text{SL}_2(\mathbb{Z})$ equivalence classes of discriminant $D$ forms and by $h(D)$ the number of $\text{SL}_2(\mathbb{Z})$ equivalence classes of discriminant $D$ primitive forms.
I want to prove that $$|\text{Cl}(D)|=\sum_{D'}h(D'),$$ where the sum is over all the discriminants $D'$ (that is, all $D'\equiv 0,1 \pmod{4}$) such that $D'|D$ and $D/D'$ is a square.
In order to gain some understanding, I did the case $D=-36$ in which $\text{Cl}(-36)=\{[1,0,9],[3,0,3],[2,2,5]\}$ and $3=|\text{Cl}(-36)|=h(-4)+h(-36)=1+2$. It is clear that $[1,0,9]$ and $[2,2,5]$ are the primitive classes of $\text{Cl}(-36)$ and that $(3,0,3)$ is a multiple of the only primitive class of $\text{Cl}(-4)$. However I can't seem to do the general case.