I've found another constraint in the form of $\displaystyle \displaystyle |f(t)|\leq M\ e^{kt} $, where $\displaystyle \displaystyle M>0, k\geq 0$

Is the one you provided stating the same as this one? What i understand from this one is that all positive values of the image of the original function must not exceed that exponential. The reason why the restriction on M is so that the exponential is only positive and the constraint on K is so that, well, if K were less than 0, then that would become $\displaystyle \displaystyle |f(t)| \leq \frac {M}{\ e^{kt}}$, which i'm guessing doesn't express the conditions of the Laplace transform. Am i correct in my judgement on that?

Why this restriction must hold, from what i have read, is so that the output of the transform is a finite value. What i do not understand is how that condition assures it will be a finite value? I am trying fairly hard to understand this and not just go by some formulas, so i would greatly appreciate further feedback

I am sorry for the late reply, but i had to look around and get myself better informed before i asked any further.