Intuitively speaking when you are given two signals/or functions $f$ and $g$. You time reverese one of the signals, it doesnt matter which one, and shift it by a value of $t$ then you simply multiply and then sum the area under the intersection.
If you consider a function say function of $x$, then time reversal means inserting $-x$ wherever you see $x$ in this function.
Example:
Question1: Assume you have a function $f(x)$ that is $1$ if $x\in[0,1]$, and $0$ elsewhere then how should you plot $f(-x)$?

Question2: Assume you have $g(x)=f(x)$ is there any intersecting area between $f(x)$ and $g(x)$?

Question3: Now shift $g(-x)$ by $0.5$, that is to find $g(-x+0.5)$. How does it look like when you plot it?

Question4: Where does the intersecting region lie in this case $x\in?$ what is the are of the intersecting region? Answer: below white area$=0.5$ at $x=0.5$ shift.

Question5: If you select the shifting parameter not $0.5$ but all reals in $[0,2]$ what function should you get at the output? check $x=0.5$ and see $f*g(x)$ is $0.5$ as found at step 4

EDIT: You define convolution integral in $[0,t]$ for bounded signals. The integral limits depend on where your signal is non-zero.
If you have two signals as you suggested $f(t)=e^{at}$ and $g(t)=e^{bt}$ then the first question: what is the relation between $a$ and $b$? are they positive? where is the function defined? For example when $a$ and $b$ are some positive terms then we have the following integral
$h(t)=\int f(\tau)g(t-\tau)d\tau= \int e^{a\tau}e^{b(t-\tau)}d\tau=e^{bt}\int e^{(a-b)\tau}d\tau=\Bigg]_{\tau\in\Omega}\frac{e^{(a-b)\tau}}{a-b}$
clearly $\Omega=\mathbb{R}$ is not possible because the integral does not converge.