Why does the frequency spectrum "fill in" under an existing envelope when you increase the period to go from Fourier Series to Fourier Transform?

I think the main problem is that people draw the "lollipops" that represent the individual values in a Fourier sum into the same diagram as a function curve, and forget to mention that this only works "visually". As soon as you'd do something like an integral over that (and that's what the inverse Fourier transform is), these isolated points would "integrate to nothing":
An integral doesn't "care" about isolated points with bounded values.

If you want to take these coefficients of a Fourier Sum and want to do similar things, you need to take them and multiply them with magic functionals – Dirac impulses – that are zero everywhere but at the position they are shifted to, and there, they don't have an actual value, but integrating over them gives the integral 1.

Now. You can't draw "a functional that's 0 everywhere but diverges at one point, at which you can only measure it by integrating over it". That's not drawable.

So people draw lollipops of some height $h$ and mean

here's a Dirac impulse, multiplied with $h$, so if you integrate across the position of the lollipop, you get value $h$ from your integral.

But: that's simply not "compatible" with the idea of "moving points closer together": because, the energy before and after transform needs to stay the same, that would mean these lollipos would need to get smaller and smaller – you're integrating more of them. Before you can make the transition from sum to integral, they'd become all zero-height.

But that's not happening. We're not "moving points together, until they become dense". (That's also impossible, mathematically: no matter how many points you add, you can always say, this is the first point, this one was added later, this one was added later, and so; you can count them. No set that is countable contains enough points to be dense in the integral sense.)

Instead, you'd follow the typical textbook logic of showing, from the other side, that for functions that are simply not square-integrable, like the humble $e^{j2\pi ft}$, the simplest of all harmonic signals, we still want to define a Fourier transform. To do that, we want that transform such that it gives value "0" for every point but at frequency $f$, and thus, when the inverse Fourier transform integrates across that point, the inverse Fourier integral must take the value $e^{j2\pi ft}$. From that, you decide you need something like the Dirac impulse, and show that the coefficients from Fourier sums are the same as the factors in front of such Dirac impulses.

Now, sadly, none of this is "intuitive" in a graphical way, because as said, these things are impossible to draw correctly. You hence will need to stop thinking as integrals as sums over numbers that get denser and denser, and start thinking of them as taking the value under the curve of a graph, and where there's no drawable curve, might take other values, such as when integrating over such a Dirac distribution. Sorry!

Sampling Theorem in Reverse

The sampling theorem states that if a continuous-time function, $x(t)$, is ideally sampled, by multiplying by a string of appropriately scaled dirac impulse functions spaced in time at discrete intervals, $T$, the spectrum, $X(f)$, of that continuous-time function is copied, repeated, and overlapped at regular intervals in the frequency domain with frequency interval $1/T$. Since the Fourier transform is an invertible (one-to-one) operator, one could restate the sampling theorem and say that if one somehow copies, repeats, and overlaps at regular intervals (of $1/T$) the spectrum of some continous time function, it would have the effect of sampling that time function at regular discrete intervals spaced by $T$.

$$ \mathscr{F}\left\{ x(t) \cdot \left( T \sum\limits_{k=-\infty}^{\infty} \delta(t-kT) \right) \right\} = \sum\limits_{n=-\infty}^{\infty} X\Big(f - n\frac1T \Big) $$

where

$$ X(f) = \int\limits_{-\infty}^{+\infty} x(t) e^{-j 2 \pi f t} \ \mathrm{d}t $$

$$ x(t) = \int\limits_{-\infty}^{+\infty} X(f) e^{+j 2 \pi f t} \ \mathrm{d}f $$

is the Fourier Transform and inverse. $\delta(t)$ is the Dirac unit impulse function.

The duality theorem of the Fourier Transform says that the roles of time $t$ and frequency $f$ can be exchanged, sometimes requiring a sign change in either $t$ or $f$. What this means is:

$$\begin{align} \mathscr{F} \Big\{ y(t) \Big\} &= \mathscr{F}\left\{ \sum\limits_{k=-\infty}^{\infty} \hat{x}(t-kP) \right\} \\ \\ &= \hat{X}(f) \cdot \left(\frac1P \sum\limits_{n=-\infty}^{\infty} \delta\Big(f -n\frac1P\Big) \right) \\ \end{align}$$

where

$$ y(t) \triangleq \sum\limits_{k=-\infty}^{\infty} \hat{x}(t-kP) $$

and

$$ \hat{X}(f) \triangleq \mathscr{F} \Big\{ \hat{x}(t) \Big\} $$

This says that copying, repeating, and overlapping a time function (call it a "wavelet" or a "grain") $\hat{x}(t)$ with repeat period of $P$ (or repeat rate or $1/P$) essentially samples the grain's spectrum at regular discrete-frequency intervals of $1/P$. This converts the continuous spectrum, $\hat{X}(f)$ of the nonrepeating grain function $\hat{x}(t)$ into a line spectrum, which is the Fourier Transform $Y(f)$ of the periodic function $y(t)$. The amplitudes of the discrete lines are the coefficients of the Fourier series of the repeated wave function $y(t)$.

The original grain spectrum $\hat{X}(f)$ can now be considered to be the "spectral envelope" of the line spectrum $Y(f)$. Note that if the repeat rate $1/P$ of the grains is changed, this changes the spacing of the line spectrum $Y(f)$, but not the spectral envelope $\hat{X}(f)$, which does not depend of P.