Confusion for two-sided signal
Given LTI system with transfer function :
$$H(s) = \frac{ -2 s}{(s+6)(s+2) } \tag{1}$$
which has the poles $s = \{ -2 , -6 \} $, and zeros $ s = 0 $.
Apply partial fraction expansion :
$$H(s) = \frac{1}{s+2} - \frac{3}{s+6} \tag{2}$$
Based on pole locations, there will be three ROCs,

ROC1 :  -$2 &lt; \mathcal{Re}\{s\} &lt; \infty $  ; stable
ROC2 : $-\infty &lt; \mathcal{Re}\{s\} &lt; -6$  ; unstable
ROC3 : $ -6 &lt; \mathcal{Re}\{s\} &lt; -2$   ; unstable

The impulse responses found from (2) and the chosen ROC , with the fact that
$$\mathcal{L}\{ e^{at}u(t) \} \longleftrightarrow \frac{1}{s-a} ~~,~~ \mathcal{Re}\{s\} &gt;\mathcal{Re}\{a\} \tag{3}$$
$$\mathcal{L}\{ -e^{at}u(-t) \} \longleftrightarrow \frac{1}{s-a} ~~,~~ \mathcal{Re}\{s\} &lt;\mathcal{Re}\{a\} \tag{4}\\\\$$

ROC1 :  -$2 &lt; \mathcal{Re}\{s\} &lt; \infty $  ;  right-sided  : $h(t) = (e^{-2t}-3 e^{-6t}) u(t)\\\\$
ROC2 : $-\infty &lt; \mathcal{Re}\{s\} &lt; -6$  ; left-sided : $h(t) = (-e^{-2t}+3 e^{-6t}) u(-t)\\\\$
ROC3 : $ -6 &lt; \mathcal{Re}\{s\} &lt; -2$   ; two-sided : $h(t) = -e^{-2t}u(-t)-3 e^{-6t}u(t)$.