exponential order
Picture two runners on a track that goes on forever. One is the function f(t) you care about; the other is a pure exponential e^(at). To say f has exponential order is to say that, sooner or later, the exponential runner is fast enough to stay ahead of f forever. As long as some exponential can outrun your function, the Laplace transform's decaying weight e^(-st) can tame it.
Formally, f(t) is of exponential order if there exist constants M > 0, a, and a time T such that |f(t)| <= M e^(at) for all t >= T. In words: past some point T, the size of f never exceeds a fixed multiple of a fixed exponential. The function may grow — even grow exponentially — but not faster than some exponential. This is the second of the two existence conditions. Here is the intuition for why it works: the transform integral contains f(t) e^(-st); if |f(t)| <= M e^(at), then |f(t) e^(-st)| <= M e^(-(s - a) t), and for s > a that bound decays fast enough that the integral converges. The threshold s > a is exactly the lower limit on s where F(s) exists.
The honest caveat hiding here: not every function passes. The classic failure is f(t) = e^(t^2). No matter how big you make a, the parabolic exponent t^2 eventually beats the linear exponent a t, so no exponential can cap it — and e^(t^2) has no Laplace transform at all. The good news is that essentially every function from a realistic model (polynomials, exponentials, sines, and their combinations) is comfortably of exponential order, so in practice the condition is met without you checking.
f(t) = t^3 is of exponential order: for any a > 0, t^3 eventually stays below e^(at), so |t^3| <= e^(at) past some T. Hence L{t^3} exists for s > 0. But f(t) = e^(t^2) is not of exponential order and has no transform.
Powers grow slower than any exponential; e^(t^2) grows faster than all of them.
Exponential order is a sufficient condition together with piecewise continuity — it guarantees the transform exists, but it is not a strict necessity for every function. Still, it covers everything you will meet, so treat it as the working rule.