As Jer pointed out, there are two ways for the belt to transverse a pulley.  I will refer to these as regular and twisted pulleys.  With these two types of pulleys there are three different ways for the belt to travel between a pair of pulleys: external when both pulleys are regular, internal when both pulleys are twisted, and crossing when one pulley is regular and the other is twisted.

The centers of the pulleys form a triangle which I will call the reference triangle.

I noticed that no matter what the configuration is there is one point on a pulley that is always in contact: the point where the external angle bisector of the reference triangle intersects the circumference of the pulley.  Similarly, there is a point that is never in contact: the point where the internal bisector intersects the circumference of the pulley.  These two points are the endpoints of a diameter of the pulley.

Then I can break down the belt into segments defined by the external bisector points. Each segment consists of two halves of a pair of consecutive pulleys and the straight belt segment between the pulleys.  Let D be the length of the side of the reference triangle and let T and U be the measures of the two angles at either end of the side. 

An external segment has its straight segment equal to D and the amount of belt wrapped around the pulleys as (pi/2 - T/2) + (pi/2 - U/2).

An internal segment is similar.  Its straight segment length is equal to D and the amount of belt wrapped around the pulleys is (pi/2 + T/2) + (pi/2 + U/2).

A crossing segment is a little more complicated.  In this case I assume T is at a regular pulley and U is at a twisted pulley.  Its straight segment length equals sqrt(D^2-4) and the amount of belt wrapped around the pulleys is (pi/2 - T/2) + (pi/2 + U/2) + 2*arcsin(2/D).

Putting the belt back together allows for some simplification.  For each regular pulley add (pi-T).  For each twisted pulley add (pi+T).  For each external and internal segment add D.  For each crossing segment add sqrt(D^2-4) + 2*arcsin(2/D).

The 6-9-13 triangle given in the problem has angles of 0.6520 (opposite side 9), 2.0731 (opposite side 13), and 0.4165 radians (opposite side 6). For Jer's arrangement, the pulleys at angles 2.0731, and 0.4165 are regular, the pulley at angle 0.6520 is twisted, the segments at sides 6 and 13 are crossing, and the segment at side 9 is external.

Then Jer's belt has a length of (pi-0.4165) + (pi-2.0731) + (pi+0.6520) + (sqrt(6^2-4)+2*arcsin(2/6)) + (sqrt(13^2-4)+2*arcsin(2/13)) + 9 = 36.0779