On the orbits and patterns of this process I started with $9$ and $5$ below and chose the first $2$ factors that popped into my head, which I then split off into two strands as $3$ and $3$. I did the same for $5.$ 
My process is, when there are two outgoing arrows from a node, factor (into $2$ factors), and when there are two incoming arrows to a node, add. 
At the end I arrived at the new numbers $9$ and $4.$ So now $9$ and $4$ are the new starting numbers, and the process continues.

Q: Are there any closed orbits? (you eventually recover the two initial starting numbers after a certain number of iterations). Do the successive numbers generally grow larger as you keep doing this process, or do they grow smaller, or neither?


 A: This answer assumes I have free choice of how to split a number into factors and where to send each factor.
First notice that in your diagram, the top half and bottom half have the same structure (graph), so it suffices to consider only one half, which looks like a "butterfly".

Are there any closed orbits? 

Yes there is at least one closed orbit.  Consider the following "butterfly" diagram - sorry about my poor "graphics"!  We start (on the left) with a pair of even numbers $(a,a)$, split each as ${a\over 2} \times 2$, send the $2$ to the middle node making it a $4$, split that as $2 \times 2$, and end up (on the right) with a pair $(b,b)$ where $b=2 + {a \over 2}$.
a -(a/2)-> b = 2 + a/2
 \       ^
 (2)    /
   \  (2)
    v /
     4
    ^ \
   /  (2)
 (2)    \
 /       v
a -(a/2)-> b = 2 + a/2

If you start with $a=4$ you get $b= 2 + {a \over 2} = 4$ again.

Do the successive numbers generally grow larger as you keep doing this process, or do they grow smaller, or neither?

This will mainly depend on how you split them into factors.  On one hand, if you split $x = x \times 1$ then the numbers will surely grow, and doing this every time it's easy to make a chain that grows forever.  
OTOH if you split $x$ in almost any proper way (i.e. not as $x \times 1$) then the numbers will usually shrink, and if you split them as evenly as possible, i.e. $x = y \times z$ where each $y,z \approx \sqrt{x}$ the numbers will shrink very quickly.
Obviously no chain can shrink forever, but you can make a shrinking chain for any length you want.  If you rename $a = f_{n+1}, b=f_n$ in the butterfly diagram above, and rearrange a bit, we have:
$$f_{n+1} = 2 ( f_n - 2)$$
So starting with e.g $f_0 = 5$ you can keep extending the chain to the left:
$$\dots \rightarrow (68,68) \rightarrow (36,36) \rightarrow (20,20) \rightarrow (12,12) \rightarrow (8,8) \rightarrow (6,6) \rightarrow (5,5)$$
