Is multiplication in physics purely mathematical or is there a physical explanation to it? Here is what I mean: We always use mathematics in physics, which is pretty powerful, but I still need to ask whether multiplication in physics has a better explanation to what I already think.
For every other operation, i.e addition, subtraction and division, I have a very intuitive situation to explain what these mean. Suppose we have two objects, having masses $m1$ and $m2$. What if we add these, $m1 + m2$? We get a new object having this mass, which clearly shows what addition means. A same explanation can be used for subtraction. As for division, we have velocity, which just means the rate of motion, and every other 'rate' quantity is pretty easy to understand.
But here is the problem: How do we explain a product, for example $W = F . s$? Is there any meaning behind this. For example, I can say that "Work is defined as the sum of a Force distance times". But what does this even mean? Do we just define quantities involving product to just calculate stuff, or just because of a conservation law(like the case in $p$), or is there any other reason behind it?
I hope I got myself cleared. 
 A: Products are omnipresent in physics – and even in less quantitative sciences – from Day One. The product $A=XY$ of two numbers may be visualized as the area $A$ of the rectangle whose sides are the two factors $X$ and $Y$. When embedded in physics, this simple formula $A=XY$ already includes units: the sides are in meters and the area is in squared meters. You may divide the rectangle to the unit squares. That's probably how children learn products when they are 7 or 8.

The height of the rectangle is six units (imagine they are meters), the width is five units, and it contains thirty squares – units of area (thirty squared meters).
But the optimum interpretation depends on the precise quantity. The way to think about every product depends on the quantities we multiply.
Take $W=\vec F\cdot \vec s$. It's the work that the force $\vec F$ does when it moves the object by the distance $\vec s$ (including directions). Why is the formula a product? Because the work is the change of some energy and the following argument works for energy. In the following argument, we move objects and we will do so in a clever way that makes it obvious how much energy we spend (how much work we do). But because the energy is conserved, the work we do must be independent of the precise way how we moved the objects.
Imagine that you have a piece of metal that is attracted to the earth by the force 1 newton (the mass is about 0.1 kilograms). To vertically lift this object by 1 meter (increase the altitude) requires us to use and fully discharge a battery of a particular size. It's always the same and let's just call the energy contained in the battery "1 newton meter" or, more precisely, "1 joule". It is just a name. If you are inconvenient with the name, call it "one battery of mine" and we will return to this naming issue.
Now, the question is how many of these batteries one needs to buy and fully "consume" for a metallic object of weight (the force) $F=5\,{\rm N}$, five newtons, to be lifted by $s=6\,{\rm m}$, 6 meters. Equivalently, what work do these batteries do?
It's easy to answer. We may cut the object to 5 smaller ones. Each of them is attracted by the force 1 newton, as previously. And we may lift these parts separately. We may lift the first piece by one meter. Then another meter... Six times. We lift it by 6 meters. We consume 6 batteries.
Then we do the same with the second object, extra 6 batteries are gone. Then with the third piece, 6 batteries gone. Fourth piece, 6 batteries. Fifth piece, 6 batteries. In total, we will consume
$$ 6+6+6+6+6 = 6\times 5 = 30$$
thirty batteries. Each of them was said to have the energy 1 joule. So we will need 30 joules. Note that the numerical part of the calculation is obvious and the picture above (which computed areas) may also show how many batteries we need. Each column corresponds to one fifth of the metallic object; each row corresponds to one meter of the distance by which the objects have to be lifted. The only "controversial" thing about the calculation is that I called the energy in one battery "one joule" which is equal to "one newton-meter".
But this has the advantage that the formula
$$ W = F\cdot s $$
whose numerical part was fine (it's clear why we got a product, right?) will work including units. If we just use units for $W$ that are naturally "products" of the units for $F$ and for $s$, then we may multiply the numbers as well as the units and both of them come out correctly. So if we want the formula to work including the units, we see that the unit of energy stored in the battery I started with should be said to be the product of one meter and one newton – this product is also called one joule.
So physicists aren't afraid of multiplying quantities with units. They routinely multiply their powers, too. For example, the accelerated motion with acceleration $a$ changes the distance by $s=at^2/2$ after time $t$. The distance is the average speed times time and the average speed is $\vec v = at/2$ (zero at the beginning, $at$ at the end).
A: If we take your same two masses and write $m_1 = s \cdot m_2$, it's easy to understand what the scalar multiplication is doing: it's scaling. $m_1$ is $s$-times as big as $m_2$. In your language, multiplying $m_2$ by $s$ gives us a new mass which has the mass of $s$ [many] $m_2$'s.
This may not seem to answer your question about work, because now the things we're multiplying have units. But we can always work in natural units where everything is dimensionless, and then the analogy holds. $\mathbf{W} = \mathbf{F}\cdot\mathbf{s}$ means that the the work is $\mathbf{s}$ times bigger than the applied force.
A: Multiplication is physical. It measures area and area is physical. It's one of the earliest physical observations that became encoded into mathematics as geometry - measurement of the earth, that is measurement of area.
A: It's important not to confuse physics with mathematics. The physics of a situation lies in the assumptions that go into a calculation, the definitions of important quantities, and the best way to frame results to gain insight about a system. The mathematics provides the tools and machinery that let us move from assumptions to results.
Multiplication is just a mathematical step. There is no intrinsic physical meaning to it. It simply takes two quantities that we previously thought were interesting and produces another quantity. The physics lies in asking questions like: what was the importance of the original two quantities, and why was it useful to multiply them together?
Specifically in the case of work, the physics lies in why we find it useful to give a name to the mathematical combination $Fs$. What motivates assigning a special meaning to this combination by giving it a name? The answer is that work has special, useful properties. For example, this definition of work allows us to prove the work-energy theory, which says that the net work done on a system is equal to the change in kinetic energy.
A: I would think most (if not all) physics equations would have a physical interpretation. BUT we can't get away from the underlying math. In your example above $W = F \cdot s$, it's multiplication  so that means just repeated additions right? That math part we can't reduce any further. I think the physics comes in from thinking about what's being repeatedly added. My short answer: changes in kinetic energy.
IF you accept that work is simply the energy needed to move an object, then we can start there. Assuming we have some mass $m$ that's not moving, it's speed is of course zero. To get it to move somewhere, we'll need to change it's speed. Let's think of flicking a marble on surface. The marble will go from $v=0$ to $v=something$ back down $v=0$. Then there is a $\frac{dv}{dt}$ that naturally comes out. Now let's say you wanted to flick a marble that was twice as big, well you'd have to hit it twice as hard for it to go the same distance. So you've already got something like "doing this work is $m \cdot \frac{dv}{dt}$ (of course this is just $F$). And this bit of multiplication, mass times change in speed, is akin to the example some previously gave with the current and the coils. There's just $n$ coils each carry $I$ current so if you want the total you just count the coils and add or you multiply, $n I$.
Let's say you do all this and the marble travels a distance $s$. Great, maybe you're done and happy. That's all the work you need to do today. BUT if you need it to go $2s$ or $3s$, and the marble is back to speed of zero, then you've got to do the whole process again and give it another 2 or 3 kicks. Then you do $m \cdot \frac{dv}{dt}$  + $m \cdot \frac{dv}{dt}$  + $m \cdot \frac{dv}{dt}$ + ...
Of course I've assumed you've been hitting the marble in the same direction as where you want it to move, but if not that's why you need the vector product and not just a simple scalar multiplication.
