The general intuition behind scalars I've recently noticed there's a small bit of a gap in my intuition of something that is very key to science and mathematics in general -- the intuition of scalars behind the fact that they should belong in formulae due to derivation. Let me try and make this more clear by considering the velocity formula for constant acceleration (0 for this case), basic stuff but highlights my point:
$$ v = \frac{d}{t}$$
Thus:
$$ vt = d$$
Let's let me set my velocity to, say, $v = 2\frac{meters}{second}$ first. Now, for each $t$, obviously we multiply the value of $t$ to the value of $v$ to find the distance traveled. So, at $t = 3s$, we find the distance traveled by adding $v$ by itself 3 times, in this case $2\frac{meters}{second}$. 
The key here is we add $v$ by itself $t$ number of times. 
So, if $v$ instead equals $12\frac{meters}{second}$, at $t = 3s$ we add 12 by itself 3 times. 
This is clearly all trivial, and I don't blame you if you're wondering how such a question can spawn in a pond so shallow, but I feel like I can't properly intuitively explain why we add $v$ by itself $t$ number of times to find $d$. That sounds a bit abnormal to me. Obviously, I can explain why this is by pointing to the velocity formula above and performing basic algebra, and that this must be the case because by contradiction, we simply could not make accurate measurements, but I don't think this is a good enough explanation. The idea of scalars is fine up until that, which embarrasses me slightly to even admit that this is not clear to me. 
I suppose my question is this: 


*

*The following statement is true: We add $v$ by itself $t$ number of times, since $t$ is a scalar with independent $v$. 

*Why is this intuitively true, other than the fact it can shown using basic algebraic manipulation, that by contradiction measurements couldn't be accurate, and that this is shown to be true by experimentation? I feel like there's some fundamental thing about multiplication hidden here that has slipped my mind. 

 A: It is not entirely clear where the confusion lies and what the question is. The following is mostly a guess based on this particular fragment.

The key here is we add $v$ by itself $t$ number of times. 

Visualizing multiplication as a repeated addition may work as a basic intuition, but even then it wouldn't make much sense to say that, for example,  $\,\sqrt{2} \cdot \pi\,$ means $\sqrt{2}$ added to itself $\,\pi\,$ times.
In the question here, however, the numbers are nice integers, and the problem lies with the proper handling of units, instead.

So, if $v$ instead equals $12\frac{meters}{second}$, at $t = 3s$ we add 12 by itself 3 times.

If we add $v$ to itself $3$ times, then we get:
$$
12\,\frac{m}{s}+12\,\frac{m}{s}+12\,\frac{m}{s}=36\,\frac{m}{s}
$$
That would be the speed of a point moving at $3$ times the speed of the first one, but not a distance.
The distance covered in $\,3s\,$ would be, instead:
$$\require{cancel}
12\,\frac{m}{s} \cdot 3\,s = 12\,\frac{m}{s}\big(1\,s + 1\,s + 1\,s\big)=12\,\frac{m}{\bcancel{s}} \cdot 1\,\bcancel{s} + 12\,\frac{m}{\bcancel{s}} \cdot 1\,\bcancel{s} + 12\,\frac{m}{\bcancel{s}} \cdot 1\,\bcancel{s} = 36 \,m
$$
A: The idea here is to see the velocity as the distance body covers in 1 second. Suppose the velocity is 2 meters per second . It means that the body is covering 2 meters every second. Therefore in 2 seconds it will cover ( 2+2 ) meters and in t seconds it will cover ( 2+2+...t times) meters.
Its essentially that when we mix intuition and algebra the result is confusion. Just don't mess yourself with units and consider the intuitive rationale behind the concepts.
Hope it helps.
