If I add a 2 ft border around, say a octagon, would the resulting side length be +2 feet or +4 feet?

Your question can be interpreted in many different ways.  Each one will give you a slightly different answer.

I have looked at just one of these.  I decided to consider a regular octagon.  The border is a concentric octagon where the width of 2 feet is at the corners.  the width between the sides would actually be a little less than this.  If you wanted the perpendicular width of the border to be the same all around then the outside shape would not be an octagon.

now, triangles OAB and triangle OQP are similar figures so their sides must be in the same ratio.

I have determined the length of QP to be  $$L_2=\frac{L_1(r+2)}{r}$$

Now  $$L_1$$   is be worked out as a function of r (original radius)  I think CPhill may have done this for you.

BUT   $$L_2$$    is still going to be dependant on the original radius of the octagon.  So I do not think that there is any set answer to you questions.  

It is an interesting question though.  

It would vary with the figure. If you add a 2 foot border around a square for instance, each side would be extended by 4 feet. For a triangle, it is more than 2 feet added to each side (but I haven't done the calculation).

The side length - s - of an octagon with radius of r feet  can be found using the Law of Sines......we have

r feet/sin 67.5  = s / sin 45

(r feet) sin 45 / sin 67.5  = s 

s = about (.765 r) feet

Now, let's suppose that the radius was extended by two feet, so that a border of two feet is added to the original octagon....so we have

(r + 2) feet / sin 67.5  = s / sin 45

(r + 2)feet sin 45 / sin 67.5 = s

s = ( r + 2)feet *(.765)  = (.765r + 1.53) feet

So...it appears that adding a 2 foot border would increase the side length by about 1.5 ft......