I am not sure about intuition in general, but regarding the step-function FT being a sync function:
Note that the shape will remain the same, but the frequencies over which the FT of a particular step-function resides is a function of the pulse-width of the original signal. Namely, expanding a function in the time-domain actually shrinks the corresponding frequency-domain function (think slowing down voice recordings, the sound gets very low i.e. lower frequency).
That being said, as you decrease the pulse-width of a particular step-function the frequency components of that signal increase because now there is more change happening (to use a loose descriptor) in a shorter amount of time.
In contrast, if we expand the step-function in the time-domain to have a longer pulse-width then there is less change and the corresponding frequency components must be much lower.
In general, I look at a function and try and get a feel for how quickly it might be changing to get a rough idea. But as I said, I don't know of any general rule of thumb here.
You are right, \$s=j\omega\$, so
$$ \omega = \frac{s}{j}= \frac{js}{j^2} = -j s $$
Substituting that into your transfer function
$$ H(j\omega) = \frac{10+j\omega}{4-\omega^2+4j\omega} = \frac{10+s}{4-\left(\frac{s}{j}\right)^2+4s} = \frac{10+s}{4+4s+s^2},$$
since \$\frac{1}{j^2} = \frac{1}{-1}=-1\$.
In order to find \$h(t)\$, you need to calculate
$$ h(t) = L^{-1}\{H(s)\} = \frac{1}{2\pi j}\lim_{T\to\infty}\int_{s=-\gamma-jT}^{\gamma+jT}H(s)e^{st}\; ds.$$
You can do this by using Cauchy's residue theorem, or else make it easier for yourself and use tables. In this case,
$$ 4+4s+s^2 = (s+2)^2, $$
so with \$s'=s+2=s-(-2)\$,
$$ H(s) = \frac{s'+8}{(s')^2}=\frac{1}{s'}+\frac{8}{(s')^2},$$
and the rest should be trivial.
Update
We have
$$\begin{align}
L\{ a f(t) + b g(t) \} &= a\,F(s)+b\,G(s) & \text{(linearity)}\\
L\{1\} &= \frac{1}{s} & \text{(constant)}\\
L\{t\} &= \frac{1}{s^2} & \text{(first order)}\\
L\{t^n\} &= \frac{n!}{s^n} & \text{($n$-th order)}\\
L\{e^{kt}f(t)\} &= F(s-k) & \text{($s$-plane shift)}
\end{align}
$$
so setting \$F(s) = s^{-1}\$, then \$f(t)=1\$ and
$$ L^{-1} \left\{ \frac{1}{s-(-2)} \right\} = e^{-2t},$$
and setting \$F(s) = s^{-2}\$ for the next term, then \$f(t)=t\$ and
$$ L^{-1} \left\{ \frac{8}{(s-(-2))^2} \right\} = 8t\,e^{-2t},$$
so you end up with
$$ h(t) = L^{-1}\{H(s)\} = (1+8t)e^{-2t} $$
Best Answer
It's important to use the right sort of tool for the job. A hammer is defined to put in nails. If you want to turn screws, use a screwdriver.
The Discrete Fourier Transform (DFT), of which the Fast Fourier Transform (FFT) calculates exactly the same result, but far faster, is defined only over repetitive signals. That is, the result it produces is exactly the same (with appropriate re-labelling) as if you had taken the signal input to the FFT, and concatenated it with itself.
This cyclic feature means that the FFT is a natural fit for repetitive waveforms. If the input sample contains an exact integer number of cycles, then it can be extended indefinitely by repetition, and look exactly the same as the original.
The technique of windowing a signal before FFT is used to reduce the problems that occur when it's used to transform a non-cyclic waveform. For example, take a one-second sample of a continuous 10.5Hz signal. If this sample is concatenated with itself, you get a half-cycle step, which messes with what you would hope to see doing a spectrum analysis. Note that it doesn't make the transform wrong, it just means that what you got is not what you naively hoped to get. If you hit the nail sideways, you get a bent rather than a driven nail, the hammer is still working just fine, doing exactly what it's defined to do.
The window reduces the signal amplitude at the beginning and end of the sample to zero, which means it can be cyclically extended. The window changes the signal, so the spectrum will change. However, it changes in a known and easily calculable way, which the FFT doesn't further change.
When used to transform a transient signal, we need to ensure that the signal sample starts and ends at zero. Then the sample meets the cyclic requirement, and the FFT will give you exactly what you would expect if you had a train of transient signals, repeating every sample length.
That's what the FFT does. If you don't want the transform of a train of signals, then the FFT is not the hammer to use. You still have a lot of freedom to adjust this train of signals to your taste, just as you can choose what window to use when transforming a continuous signal. You can make the impulses repeat less often by increasing the time length of the sample for instance.