Finding / Proving the Order of Dihedral and Symmetric Subgroups Lately I've been trying to follow along with this YouTube lecture series called "Visual Group Theory" by Clemson's Mathew Macauley and have had some issues grasping the concept of a group's "order".. 
I (think I) understand how it works when referring to reflections, rotations, etc. in the sense that we are looking for the $"k"$ value that makes $x^k = e$ 
so examples when looking at $D_4$ where $R = 90^\circ rotation$ and $F = horizontal flip$ :


*

*$|R^2| = 2$  since two rotations bring you halfway and so doing this twice will preserve it's footprint

*$|RF| = 2$ since when performing this on a piece of paper with numbered corners i got back to my original starting point after performing RF twice


but when moving into the proof of orders that are dealing with numbered groups i have trouble wrapping my head around what's going on.. 
Proposed with an exercise of examining $S_4$, when computing $|(12)(13)|$ or $|(1243)|$ i lack the intuition of seeing how this gets mapped and how we deduce the order of it.. (although my guess would be $|(12)(13)| = 4$ since it can be rewritten as $|(123)|$)
please let me know what you think; any and all help, input, and assistance is greatly appreciated!

You started with a slight error.  There's a difference between the order of a group, and the order of an element.  To illustrate, a favorite theorem of mine is Cauchy's theorem. 
 It states that if $p$ is a prime dividing the order of the group, then the group has an element of order $p$.  
But getting back to your question, actually $|(12)(13)|=|(123)|=3$, because a $3$-cycle has order $3$.  More generally, an $n$-cycle has order $n$. Your intuition will develop as you move along.
Here's a useful fact: if $a,b\in S_n$ are disjoint, then $|ab|=\operatorname{lcm}(|a|,|b|)$.
So for instance, let's look at $(12)(34)$.  Since the transpositions $(12)$ and $(34)$ are disjoint, the order of their product is $2$.  To contrast, in your example $(12)(13)$, the transpositions $(12)$ and $(13)$ are not disjoint, as they both "move" $1$.
That's all for now.
