Intuition about how Voronoi formulas change lengths of sums

In reading the literature one encounters countless examples of Voronoi formulas, i.e., formulas that take a sum over Fourier coefficients, twisted by some character, and controlled by some suitable test function, and spits out a different sum over the same Fourier coefficients, twisted by some different characters, and this time controlled by some integral transform of the test function.

The reason one wants to do this in practice is that the second sum is somehow better, of course, which in my (admittedly limited) experience tends to boil down to the length of the second sum having changed significantly to the better.

I'll give an example (from Xiaoqing Li's Bounds for GL(3)×GL(2) L-functions and GL(3) L-functions, because it is what I happen to have in front of me). In this case we have the GL(3) Voronoi formula $$\sum_{n > 0} A(m, n) e\Bigl( \frac{n \bar d}{c} \Bigr) \psi(n) \sim \sum_{n_1 \mid c m} \sum_{n_2 > 0} \frac{A(n_2, n_1)}{n_1 n_2} S(m d, n_2; m c n_1^{-1}) \Psi \Bigl( \frac{n_2 n_1^2}{c^3 m} \Bigr),$$ where $$\psi$$ is some smooth, compactly supported test function, $$\Psi$$ as suggested above is some integral transform of it, $$A(m, n)$$ are Fourier coefficients of (in this case) an SL(3) Maass form, $$(d, c) = 1$$, and $$d \bar d \equiv 1 \pmod{c}$$.

(I've omitted lots of details here, but the details, I think, aren't relevant to my question.)

Doing so essentially transforms the $$n$$-sum into the $$n_2$$-sum, where, as is evident in the formula, the $$n_2$$-sum has a very different argument in its test function.

What happens in practice now is that, once we get to a point where applying the Voronoi formula is appropriate, we transform the sum and study the integral transform, chiefly by means of stationary phase analysis in order to find what length of the new $$n_2$$-sum is.

In the particular example at hand, this, after identifying the stationary phase and playing along, this takes us from an $$n$$-sum on $$N \leq m^2 n \leq 2 N$$, i.e., $$n \sim \frac{N}{m^2}$$, to an $$n_2$$-sum on $$\frac{2}{3} \frac{N^{1/2}}{n_1^2} \leq n_2 \leq 2 \frac{N^{1/2}}{n_1^2},$$ i.e., $$n_2 \sim \frac{N^{1/2}}{n_1^2}$$, which then means that the arguments in the test function are now of size $$\frac{N^{1/2}}{c^3 m}$$.

I can go through the motions of performing this stationary phase analysis and so on, but my question is this: is there any intuition to be had about how and in what way these Voronoi formulas alter the lengths of sums?

First of all, the description of $$\psi$$ after the first display is confusing (assuming OP meant $$\psi$$ is supported around $$N$$, otherwise conclusion form the first display does not make sense). I went to the relevant part (end of p.318) of Li's paper and found that $$\psi$$ is not just any test function, it has a weight of $$n^{-3/4}$$, and, more importantly, it has oscillation. A necessary display of the LHS would be (taking $$m=d=1$$) $$\sum_{n}A(1,n)e\left(\frac{n}{c}+2\sqrt{n}-\frac{1}{\sqrt{n}}\right)\psi(n/N),$$ where $$\psi$$ is a test function supported on $$[1,2]$$.
To have an intuition about what the length of the dual sum would be one can follow the general heuristic formula (HF) below. $$\text{length of the dual sum} = \frac{\text{total conductor}}{\text{length of the original sum}}.$$ Here total conductor is the conductor of the oscillating object taken all the twisting into account. For example, the total conductor of $$L(1/2+it,\pi'\otimes\pi)$$ where $$\pi$$ and $$\pi'$$ are fixed $$\mathrm{GL}(n)$$ and $$\mathrm{GL}(m)$$ automorphic representations, is $$t^{nm}$$.
Conductor of $$e_q(x):=e(x/q)$$ is $$q$$. In this case the denominator in the entry of $$e()$$ is of size $$c\sqrt{N}$$. But there is twist by a $$\mathrm{GL}(3)$$ Hecke eigenvalue. So the total conductor is $$c^3N^{3/2}$$. Applying the above formula one obtains the length of the dual sum.
One can check that the above HF also occurs in the formula of the approximate functional equation of the central $$L$$-value. There are two sums in the formula corresponding to the representation and its contragredient. One can check that the product of the length of the sums equals to the conductor of the $$L$$-function.