Suppose we have a cone and we wish to parallel transport a vector $w=(0,1,0)$ from along the curve $\alpha(s)=(\sqrt{2}/2 \cos(v\sqrt{2}),\sqrt{2}/2 \sin(v\sqrt{2}),\sqrt{2}/2)$ from $p=\alpha(0)$ to $q=\alpha(\frac{\pi\sqrt{2}}{4})$
I take polar coordinates on the plane and construct the isometry $F:(\rho,\theta)\in (0,\infty)\times(0,\sqrt{2}\pi) \longrightarrow F(\rho,\theta)=\left(\frac{\sqrt{2}}{2} \rho\cos(\theta\sqrt{2}),\frac{\sqrt{2}}{2} \rho\sin(\theta\sqrt{2}),\rho\frac{\sqrt{2}}{2}\right)$.
$F$ is a local isometry between the plane and the cone, so it is equivalent if we do parallel transport on the plane or on the cone. See this image from Do Carmo's Differential Geometry of Curves and Surfaces. 
So now I know the angle between the parallel transport of $w$ and the $\alpha'(\frac{\pi\sqrt{2}}{4})$.
But how do I finish? How do I get the parallel transport of $w$?
Best Answer
You compute $q$, and $v = \alpha'(\frac{\pi\sqrt{2}}{4})$, and let $v' = v/\|v\|$, then compute a second unit vector $u'$ perpendicular to $v'$, lying in the tangent plane to the cone at $q$.
Now you take $\cos \theta v' + \sin \theta u'$ (where $\theta$ is your computed angle between the transported $w$ and $v'$).
In other words, you take the $t(s)$, $n(s)$ basis for the plane from the right hand diagram and map it to the $v', u'$ basis for the cone at $q$. (Here $n(s)$ denotes a vector in the plane perpendicular to $t(s)$.)
Yes, you have to worry about the orientation of $u'$; there are two choices for $u'$ (or, if you like, for $n(s)$.