What is meant by multiplication in physics such as $s = v·t$? What is the meaning behind multiplication in physics? Is multiplication in physics purely mathematical or there is a physical explanation to it? How do we explain the product for example, $s = v·t $? Is there any meaning behind this? For example, I can say that "Distance is defined as the product of velocity 'times' time"? But what does this even mean? 
 A: Physics can be thought of as the subset of mathematics which happens to make correct predictions about the Real World (tm) [at least roughly speaking]. As such in my view it's probably a mistake to read too much into the "physical meaning of multiplication" beyond the mathematical definitions. 
Of course there are lots of friendly-sounding examples to give people of intuition about multiplication - e.g. counting groups of something. $4\times5$ can be thought of as 4 groups of 5 things each, and this remains true for most physical quantities in some sense. However, the math doesn't care about this interpretation of the expression - the math stands by itself. So it's perfectly reasonable and self-consistent to drop the fluff and focus on the cold-hard math. 
A: Perhaps the simplest way multiplication can be understood in linear relations such as in your example is: “is proportional to with this coefficient.”
Distance is proportional to time. The coefficient of proportionality is called “velocity.”
A: Many, albeit far from all, "multiplications" in physics are simplified versions of integrals. For example, in your case, saying $\Delta s = v \Delta t$ comes from the definition of velocity 
$$ \vec{v} = \frac{d\vec{s}}{dt} $$
for which we can easily see this has come from an integral:
$$ \vec{s} - \vec{s}_0 = \int_{t_0}^t \vec{v}(t)dt. $$
This equation -- as well as $\Delta s = v\Delta t$ -- tell you that the change in position of an object from $\vec s_0$ to $\vec s$ is the result of a velocity $\vec{v}$ that occurs over time period $t-t_0$. During that time period, the velocity $\vec v$ uses every infintesimal moment $dt$ to change the position by a tiny amount $d\vec s$, until all the little $d\vec s$'s have accumulated up to $\vec{s}-\vec{s}_0$.
There are numerous other places where this occurs: $F = -\frac{dU}{dx}$, $I = \frac{dq}{dt}$, $\Delta V = - \frac{d\Phi_B}{dt}$, $P = - \frac{dU}{dV}$, and many more.
Another example is Newton's second law -- $F=ma$. If only conservative forces act on your object then you can see it is equivalent to
$$ \frac{dU}{dx} = -m\frac{dv}{dt}.$$
That is to say -- the changes in potential energy over displacement (force) cause the velocity to change over time in proportion to mass. 
In the context of vectors, scalar and vector multiplication also contribute the important directions of products, e.g. in Maxwell's equations we have
$$ \nabla \cdot E = \frac{\rho}{\epsilon_0} $$
and 
$$ \nabla \times \vec{B} = \mu_0\left[\vec{J}+\epsilon_0\frac{\partial \vec{D}}{\partial t}\right]$$
The first tells you that electric field diverges from a source in a way consistent to the distribution of source charge. The second tells you that the magnetic fields produced from real or displacement currents curl around the source and form a new vector. 
All multiplication in physics has physical meaning. Often this can be found by looking at the underlying definitions of the quantities. After all, the fundamental equations in physics -- Newton's 2nd law, the Schrodinger equation, Maxwell's equations, and so forth -- are each differential equations that link the interactions and energy of/between objects to their motion through spacetime using multiplication. 
A: Basically the multiplication in physics is the same as in mathematics.
But there is one more important thing you need to keep in mind.
Unlike in mathematics, in physical formulas the values have a number and a unit.
Therefore, when multiplying two physical values,
then you need not only to multiply their numbers.
You also need to multiply their units.
Let's take for example a car  driving with speed $v = 50 \text{ miles/hour}$,
during a time $t = 2 \text{ hours}$.
Then you can calculate the distance driven by the car by
$$s = v \cdot t$$
Using the values from above and doing the multiplication you get
$$\begin{align}
s &= v \cdot t \\
  &= 50 \text{ miles/hour} \cdot 2 \text{ hours} \\
  &= (50 \cdot 2)\ (\text{miles/hour} \cdot \text{hours}) \\
  &= 100 \text{ miles}  
\end{align}$$
A: Another way of thinking about this intuitively is to first understand the meaning of speed. If a car has a speed of 60 miles per hour, this means that every hour the car traverses 60 miles. So, if 2 hours have elapsed, the car traverses 120 miles which is 60*2. Abstracting this now to variables, if a car is moving with a speed $v$, and a time $t$ has elapsed, the distance the car traverses is $v*t$.
This is precisely the same interpretation multiplication has in mathematics. If your teacher hands 3 cookies per student, and they have 30 students, how many cookies do they have to hand out? It's 30*3=90.
A: Distance covered = velocity × time
is saying that "fastness of the object" multiplied by ``time taken'' gives the total distance covered (for linear and uniform velocities only). 
The ``Physical interpretation of multiplication'' comes from the units involved. Without the physical units all operations are just pure mathematics.
To find a physical meaning in the mathematical operations like product,addition,subtraction and division would change from context to context in physics. 
A: Multiplication in physics does have meaning. For instance take the very example you gave, "what is distance?"
We know that velocity is the distance travelled in one unit time, say like one second. Now, if you travel for 10 seconds, you are essentially travelling the distance travelled in 1 second ten times. 
If numbers fit better, consider this situation. Let us assume that you can walk 5m in one minute. In 10 minutes you would have walked 50 metres, very obviously. So, here you are travelling 'time' times the 'velocity' i.e. '10 time' times the distance travelled in 'one time'. 
Notice that the factor of time cancels out and you are left with 10 times (not 'time') the distance, which is what we want.
See the image below

We can break the 5 seconds into 5 interval s of 1 second, and in each second, the body travels a distance equal to the velocity (by the definition of velocity), and added the distances.
