I'll try my hand at this also. In my experience, in courses where the concept of convolution is taught, it's common to see the "flip-and-slide" process diagrammed out as you've alluded to. I don't feel that the graphical illustration itself really gives that much insight, but here's how you might arrive at it starting with the convolution integral itself:
$$
y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau
$$
In the event that the impulse response $h(t)$ is causal (i.e. it is zero $\forall t < 0$, and it usually is), then the above integral can instead be written as:
$$
y(t) = \int_{-\infty}^{t} x(\tau)h(t-\tau)d\tau
$$
Note that you can change the limits of the integral, because $\forall \tau > t$, the argument to $h(t)$ is less than zero, therefore making the integrand zero. Likewise, if the input signal $x(t)$ is causal, then the above integrand is zero $\forall \tau < 0$, simplifying it further to:
$$
y(t) = \int_{0}^{t} x(\tau)h(t-\tau)d\tau
$$
This is starting to look a bit less imposing. Now, the question is, how does this equation lead to the "flip and slide" interpretation of convolution? Well, by straightforwardly reading the above equation, you can recognize the following:
In order to calculate the output of the convolution $y$ at some point in time $t$, we sum up a bunch of terms over the variable of integration $\tau$, ranging from $0$ to $t$.
Each term consists of the product of the impulse response evaluated at the variable of integration $\tau$ and the input signal, time-reversed with respect to the variable of integration $\tau$ and shifted to the left by the current output time $t$. This is where the concept of "flipping" the impulse response as an interpretation of the convolution integral comes from.
However, understand one thing: there is no actual time reversal of the impulse response that occurs when you excite an LTI system with an input signal. That's merely one way of intuitively interpreting the structure of the convolution equation, and one that lends itself well to a graphical explanation that most will probably have seen in an undergraduate signals and systems course. One such example is given on Wikipedia, duplicated below. It is interpreted as follows:
$\hskip1in$
When you see the impulse response "sliding" across the input signal, that illustrates the progression of the time variable $t$.
At any particular time instant instant $t$, in the animation, remember that the output signal is equal to
$$
y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau
$$
I kept the generic form here to better fit the diagram on Wikipedia, which is not causal. We take the product of the input signal $x(\tau)$ and the time-reversed, shifted impulse response $h(t-\tau)$ and integrate over the range of $\tau$ where the two functions overlap.
This action is equivalent to finding the area under the curve formed by the product $x(\tau)h(t-\tau)$. This area is shown as the yellow area in the animation.
At any time $t$, the output of the system $y(t)$ is equal to the amount of area shown in the animation at that particular time.
As I said, that's the description that you've probably gotten from your professors, and it works well if you're tasked with calculating convolution integrals for simple functions, which typically involves some variation of the above process over a few piecewise intervals. Like I said before, though, I don't feel like this interpretation gives you much intuition to go on.
In my opinion, the superior explanation is that espoused by Dilip in the linked answer. You can arrive at it via what I think is an even more straightforward reading of the convolution integral:
$$
y(t) = \int_{0}^{t} x(\tau)h(t-\tau)d\tau
$$
Read it as follows:
The time function $y(t)$ consists of the sum of a bunch (an infinite number, actually) of functions of $t$.
Each term in the sum has the form:
$$
x(\tau)h(t-\tau)
$$
This is a function of $t$, expressing a copy of the system's impulse response $h(t)$, delayed in time by $\tau$ (the variable of integration) and weighted by the input signal $x(t)$, evaluated for $t=\tau$ (again, the variable of integration.
So, intuitively, we form the output signal $y(t)$ by adding an infinite number of copies of the impulse response. Each copy of the impulse response is delayed by $\tau$ and weighted by the input signal evaluated at that value $\tau$. For the causal case, $\tau$ ranges from zero to $t$.
That's it! I feel that this gives a much better feel for the "action" that an LTI system performs to an input signal. Think of it this way (which is very mathematically imprecise): at each time instant, the input signal "causes a copy of the impulse response to start coming out of the system," with an amplitude commensurate with the value of the input at that instant. Due to the linearity and time invariance of the system, all of these impulse response copies sum up to form the composite that is the output signal $y(t)$. While I find this very intuitive, it is even more clear for the discrete-time case:
$$
y[n] = \sum_{k=0}^{n} x[k] h[n-k]
$$
This is the exact same concept, except the infinitesimally-spaced integrals turn into discrete sums (and the discrete nature makes it even easier to visualize, as Dilip tabulated it in his answer that you referenced). I think that using this idea is the best way to express the nature of the convolution integral.
