Inverse Fourier Transform of transfer functions
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
$$\begin{align} 
L\{ a f(t) + b g(t) \} &amp;= a\,F(s)+b\,G(s) &amp; \text{(linearity)}\\
L\{1\} &amp;= \frac{1}{s} &amp; \text{(constant)}\\
L\{t\} &amp;= \frac{1}{s^2} &amp; \text{(first order)}\\
L\{t^n\} &amp;= \frac{n!}{s^n} &amp; \text{($n$-th order)}\\
L\{e^{kt}f(t)\} &amp;= F(s-k) &amp;  \text{($s$-plane shift)}
\end{align}
$$
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
$$ h(t) = L^{-1}\{H(s)\} = (1+8t)e^{-2t} $$