::: You seem to have missed my point.  Rotation of a solid object in space by an angle of θ around the x-axis is represented by a quaternion q = exp(−iθ/2), and the corresponding rotation of any vector r is calculated as qrq*, where q* is the conjugate of q (negate the vector part only).  This will rotate any vector (or an object described by a number of individual vectors) by the correct amount around the x axis.  In contrast, your construction rotates only a vector oriented to the corner of the spherical triangle correctly, and the great-circle "spherical vector sum" only works for that vector, but you seem to be wanting to interpret it as a rotation for any vector.  To take your example of two 90° rotations, say around the x-axis and then around the y-axis, we'd put q=exp(−iπ/4)=(1−i)/, q′=exp(−jπ/4)=(1−j)/, giving us q′q=(1−j)(1−i)/2=(1−i−j−k)/2=exp(−((i+j+k)/)π/3), which corresponds to a rotation of 2π/3=120° about an axis equiangular from the three +x, +y and +z semi-axes.  And as you've said, the original vector in the actual rotation does not move on a great circle in the composed rotation, but the semi-angle vectors describing the corners of the spherical triangle picture of the rotations are joined by three great circle arcs of 45°, 45°, making a third side of 60°.  You must keep in mind that the triangle picture does not happen in the same space as the rotation being described; they're related by a halving of angles around the origin. 
 ::: By way of contrast, if you take the 90° spherical triangle (like in the picture, with each corner being 90°), if you want to interpret that as representing a rotation of an object, the angles of rotation are double, thus giving 180° actual rotations being represented.  Take a book, put it on the desk in front of you.  Rotate it 180° around the x-axis pointing to your right, leaving the book face down and its top towards you.  Next rotate it 180° around the y-axis parallel pointing away from you along the desk, leaving the book face up but top towards you.  This combination is equivalent to a 180° rotation around the z-axis upwards out of the desk.  This interpretation works exactly, for all angles and axes of rotation.  If you do not use the halving/doubling of angles, everything falls apart.  —