GH from MO gave in his answer a bound due to himself and Valentin Blomer of $O(\sigma(n))$. I thought it would be interesting to compute the $O$. Here I'm looking for an effective bound of the form $\leq C \sigma(n) + o(\sigma(n))$$\leq C \Delta^{-\delta} \sigma(n) + o(\sigma(n))$ where the $C$ is explicit but the little $o$ need not be, so we only need be concerned with the Eisenstein term.
The formula in the cited paper is $$\frac{\pi^2 n}{d} \prod_p \chi(p) \leq \frac{\pi^2 n}{\sqrt{d}} \cdot 4 d_0^{1/6} 2^{ \omega(d_1)} \prod_{p \geq 2} (1+p^{-2}) \frac{\sigma(n)}{n} $$
where $d$ is the discriminant, $d_0d_1=d$ and $d_1$ is prime to $d$, so $2^{\omega(d_1)}\leq (2/\sqrt{3}) d_1^{1/2}$
$$\leq \frac{8}{\sqrt{3}} \pi^2 \prod_{p \geq 2} (1+p^{-2}) \sigma(n)$$
Now $$\prod_{p \geq 2} (1+p^{-2}) = \frac{ \zeta(2)}{\zeta(4)} = \frac{15}{\pi^2} $$
so we get the explicit bound for the main term of $40 \sqrt{3} \sigma(n)$. However this is likely not sharp. Because the bound goes to $0$ with the discriminant, one could in principle write down a finite list of forms capable of ever getting above $30 \sigma(n)$, say, and try to explicitly calculate the greatest possible ratio of the main term to $\sigma(n)$ for each one.
Also this bound is not so bad. It's only this weird factor of $2/\sqrt{3}$, times another factor of $2$, above the conjectured best possible.
I'm trying to sharpen this upper bound a little to get a better handle on the situation.$$\frac{\pi^2 n}{\sqrt{\Delta/16}} \prod_p \chi(p)$$
Define $f_Q(p)$ to be the max over $n$ of $\frac{\chi(p)}{p^{v_p(d)}} (\sum_{t=0}^{v_p(n)} p^{-t}) $$\frac{\chi(p)}{p^{v_p(\Delta)/2}} (\sum_{t=0}^{v_p(n)} p^{-t}) $
Then the main term
$$\frac{\pi^2 n}{d} \prod_p \chi(p) \leq \pi^2 \sigma(n) \prod_p f_Q(p)$$$$\frac{4\pi^2 n}{\sqrt{\Delta}} \prod_p \chi(p) \leq 4\pi^2 \sigma(n) \prod_p f_Q(p)$$
So our goal is to upper bound $f_Q(p)$. According to the paper, for primes not dividing the discriminant $d$$\Delta$, $\chi(p)= \left(1- \left(\frac{d}{p} \right)p^{-2}\right) \sum_{t=0}^{v_p(n)}\left(\frac{d}{p} \right)^t p^t$$\chi(p)= \left(1- \left(\frac{\Delta}{p} \right)p^{-2}\right) \sum_{t=0}^{v_p(n)}\left(\frac{\Delta}{p} \right)^t p^t$ so $f_Q(p) = 1- \left(\frac{d}{p} \right)p^{-2}$$f_Q(p) = 1- \left(\frac{\Delta}{p} \right)p^{-2}$. So the primes not dividing the discriminant contribute $\prod_p\left(1- \left(\frac{d}{p} \right) \right)= L(\chi_{\sqrt{d}},2)^{-1}$$\prod_p\left(1- \left(\frac{\Delta}{p} p^{-2} \right) \right)= L(\chi_{\sqrt{\Delta}},2)^{-1}$. However for a crude upper bound, we instead use $\chi(p) \leq 1+1/p^2$.
For odd primes dividing the discriminant, the bounds given for $\chi(p)$ depend on whether $p$ divides $n$ or not. If $p$ does not divide $n$, they depend further on $n_1$, which is the rank of the quadratic form mod $p$. In this case $\chi(p) \leq 2$ if the rank is $1$ but is at most $1+1/p$ otherwise. The rank can only be $1$ if $v_p(d) \geq 3$, because each term in the formula for the determinant $d$ of the symmetric matrix will be divisible by $p^3$. If $p$ does not divide $n$, the bound for $\chi(p)$$f_Q(p)$ is $p^{v_p(d)/6} (1+p^{-2}) (1+p^{-1})$. So we have
$$ f_Q(p) \leq \max \left( \frac{1 + 1/p}{p^{v_p(d)/2}}, \frac{1 + 1_{v_p(d) \geq 3}}{p^{v_p(d)/2}}, \frac{ 1 + p^{-2}}{p^{v_p(d)/3}}\right)$$
One can see in particular that all these terms areThe third contribution is always the greatest as long as $<1$$(1+1/p) \leq p^{1/6} (1+ 1/p^2)$, which happens for all $p>3$$p>2$, and in fact go to $0$ with $p$$2 \leq \sqrt{p} (1+1/p^2)$, so these local contributions go towhich happens for $0$ as the discriminant goes to$p>3$ and only fails by a factor of $\infty$$.962$ for $p=3$.
The only issueremaining contribution is the local contribution at the prime $2$. The paper gives the bound $4 \cdot 2^{ (v_2(\Delta)-4)/6}$ for $\chi(2)$ and hence $f_Q(2) \leq 2^{4/3} 2^{-v_2(\Delta)/3}$.
Hence $$\prod_p f_Q(p) \leq \frac{2^{4/3}}{(1+1/4)}\frac{2}{\sqrt{3} (1+ 1/9)}\frac{1}{\Delta^{1/3}} \prod_p (1+1/p^2)$$
Using
$$\prod_p (1+ p^{-2}) = \frac{\zeta(2)}{\zeta(4)}=\frac{15}{\pi^2}$$ we get an upper bound for the main term of
$$ \frac{2^{4/3}}{(1+1/4)}\frac{2}{\sqrt{3} (1+ 1/9)}\frac{60 \sigma(n)}{\Delta^{1/3}}= 125.697\dots \frac{\sigma(n)}{\Delta^{1/3}}$$
To get the ratio under $30$, whichthen, we need $\Delta>73$.
Combined with Valentin and GH's arguments, I believe this implies that there are only finitely many counterexamples to "the number of representatives is a huge messat most $30 \sigma(n)$" with $\Delta>73$.
It might be possible to prove a sharp bound by:
bounding the error term explicitly to eliminate large $\Delta$ counterexamples.
Explicitly calculating the main term for medium $\Delta$, instead of using this crude bound, to eliminate medium $\Delta$ counterexamples.
Explicitly calculating the main term and showing the error term vanishes for small $\Delta$, explicitly calculating the highest examples.
