This is not an answer and I cannot comment due to lack of reputation, but it seems to first pop up in his Lemma 4 on p. 50. This occurs after a long, very technical discussion about double coset operators. The goal is to show that two specific representations of the Hecke algebra are equivalent (one whose image lies in $End_{\mathbb{C}}(S_{k+1/2}(N, \chi))$ and the other in $End_{\mathbb{C}}(S_{2k}(N))$). The Lemma:

**Lemma 4 (p. 50).** Every elliptic or hyperbolic conjugacy class in $\Gamma_{0}(4N)C_{n}\Gamma_{0}(4N)$ contains an element $\begin{pmatrix} a 
& b \\ c & d \end{pmatrix}$ with $d > 0$, $(b, d) = 1$, and $\left( \frac{b}{f}, \frac{t^{2} - 64n^{2}}{f^{2}} \right) = 1$. Here, $t := a + d$ and $f := (d-a, b, c)$. 

Now, for *squarefree* $M$ set

$$\mu(t, f, n, M) := \prod_{p \mid (M, f)} (1+p)\cdot \#\{x \in \mathbb{Z} : 1 \leq x \leq M, (x, M) = 1, x^{2} - tx + n \equiv 0 \mod (fM, M^{2})\},$$

**Lemma 5.** Let $A \in GL_{2}(\mathbb{Z})$ be an elliptic or hyperbolic matrix with $\det A = 16n^{2}$, $t \equiv 0 \mod{4}$ and $f$ odd. Then *for $N$ odd and squarefree* there are $\mu(\frac{t}{4}, f, n^{2}, N)$ matrices $B \in \Gamma(1)/\Gamma_{0}(4N)$ with $B^{-1}AB \in \Gamma_{0}(4N)C_{n}\Gamma_{0}(4N)$. 

So it seems that the definition of $\mu$ depends on $M$ being squarefree, and in turn this allows Kohnen the counting argument in Lemma 5. I don't know why squarefree-ness is needed from this, though. But maybe this information can help you or someone else determine the answer.