Electronic – How to obtain 3db frequency from transfer function

The impulse response \$h(t)\$ is the time derivative of the step response. Thus, for the step response given

$$h(t) = \frac{d}{dt}S(t) = 4e^{-4t}u(t) + (1 - e^{-4t})\delta (t) = 4e^{-4t}u(t)$$

Then, the transfer function \$H(s)\$ is the transform of the impulse response which is as you alternately derived,

$$H(s) = \frac{4}{s + 4}$$

Finally we have

$$y(t) = h(t) \star x(t) = \int_0^t h(\tau)x(t - \tau) d\tau = \int_0^t x(\tau)h(t - \tau) d\tau$$

and

$$Y(s) = H(s)X(s) $$

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 $$ h(t) = L^{-1}\{H(s)\} = (1+8t)e^{-2t} $$