Hint: You can write this using Heaviside Unit Step functions (plot this versus your piecewise function) as:
$$f(t) = t^2(u_0(t) - u_{3}(t)) + 9(u_{3}(t))$$
The Laplace Transform of this is:
$$\mathscr{L} (f(t)) = \dfrac{2}{s^3}-\dfrac{e^{-3 s} (9 s^2+6 s+2)}{s^3}+\dfrac{9 e^{-3 s}}{s}$$
The Laplace transform is defined as:
$F(s)=\int _0 ^{+\infty} e ^{-st} f(t) dt$
Your first question: As one can see the limit of the integral is from $0$ to $\infty$. So, it is inherently assumed that $f(t)$ is zero for $t<0$. As a result, when we talk about $f(t)=t$, it is actually $f(t)=t, t\geq0$, which is a piecewise continuous function.
Second question: A function $f(t)$ is said of exponential order if there exists a constant $a$ and positive constants $t_0$ and $M$ such that $|f(t)|<Me^{at}$, for all $t>t_0$ at which $f(t)$ is defined. This condition should hold because otherwise the Laplace transform will not exist. To prove it, lets first split the integral into two parts:
$F(s)=\int _0 ^{+\infty} e ^{-st} f(t) dt = \int _0 ^{t_0} e ^{-st} f(t) dt+\int _{t_0} ^{\infty} e ^{-st} f(t) dt$
It is easy to show that the first part exists. Now, we need to show that the existence of the laplace transform depends on the convergence of the second part:
$|\int _{t_0} ^{\infty} e ^{-st} f(t) dt|\leq \int _{t_0} ^{\infty} |e ^{-st} f(t)| dt \leq \int _{t_0} ^{\infty} Me ^{-st} e^{at} dt = M\int _{t_0} ^{\infty} e^{-(s-a)t} dt$
This integral converges if $s>a$. So, we conclude that if $f(x)$ is exponentially ordered, there exists a constant $a$ in which $F(s)$ exists for $s>a$.
Best Answer
We have
$$G(s) = \frac{e^{-2s}}{s^2}$$
We know that
$$\mathcal{L}^{-1}\left ( \frac{1}{s^2}\right) = tu(t)$$
and that
$$\mathcal{L}^{-1}\left ( e^{-as}F(s)\right) = f(t-a)u(t-a) \tag{time shift property}$$
Therefore,
$$\mathcal{L}^{-1}\left ( G(s)\right) = (t-2)u(t-2) = \begin{cases} 0 & t \leq 2 \\ t-2 & t > 2 \end{cases}$$