We're going to call the horizontal edge $e$, the front two edges $a$ and $b$, and the back two legs $c$ and $d$. Also, we see that the numbers 11, 10 and 9 must all be in feet, as they're too big to be anywhere else.

Let's try $S=14$ first. We may as well take $a=3$, as the legs are all interchangeable at this stage.

Now if $b=4$, then we have $c=5$ (as the back legs are interchangeable). Also, we must have $d$ and the back foot as 6 and 8(in some order), and we soon see that this isn't possible (as we'd also need $e$ plus the top left hip to add up to 7). If $b$ isn't 4, then we can assume that $c=4$. But similarly we find that we can't have $b=9$ or $b=8$ or $b=7$. So $S$ isn't 14.

Let's try $S=16$. But then we need four pairs of numbers that sum to 16 (one for each leg). We try: (11,5), (10,6), (9,7), (8,?)and we can't repeat the 8 so that isn't possible.

So let's try $S=15$. This time, we may as well take $a=4$. There are only 4 pairs that add up to 15: (11,4), (10,5), (9,6) and (8,7). Call the number in the top left hip $f$.

If $b=5$, then we have $c=6$ and $e+f=6$, which isn't possible (as we don't have the right numbers left).

So we can take $c=5$. If $b=8$, have $e+f=3$ and then we'll need $f+g=4$, and this isn't possible. If $b=7$, have $e+f=4$ and we also have $f+g=4$ and this isn't possible either. Also we can't have $d=8$, as this is too big. So the only possibility is the one shown below (and this does indeed work).

(When I say the only possibility, of course the front legs could be switched, or the back legs, or the front and back swapped, but it's really the same caterpillar.)