If two points $P,Q$ which are not umbilical are given, then there are in general several geodesics but only one that is shortest among them all. Their  $ u^{'}, v^{'}$ are different, but unique for any one of the geodesic choices.

If $P,u^{'}, v^{'}$ are given there is no guarantee that the line so defined passes through $Q.$ 

These can be physically verified between two points on a cylinder/cone using a taut thread for example. 

EDIT1:

It can be compared to the dynamic geodesic trajectory situation when a gun is fired with a given velocity and inclination from start point $P$. The initial velocity and direction should be adjusted to make trajectory pass through another desired point $Q$ in the vertical plane, as it cannot meet a desired point with arbitrary setting of dynamic parameters like $ u^{'}, v^{'} $ in case of geometrical geodesics.

EDIT2:

The solution also reflects in numerical procedures adopted. We have both initial value and boundary value problems. The former is straight forward and for the latter position/ derivative values at $P$ and $Q$ are given as input and numerical iterative (again!) shoot-through technique is employed.

EDIT3:

Actually the answer to your question is determination of the common geodesic invariant through  $P,Q$. Geodesic curvature $k_g$  vanishes for all geodesics, so condition for them to be on the same geodesic line is that they be on same filament of the tangent fibre bundle  sharing the same invariant.
The parameters of the *two points* go to determine it..

For a surface of revolution the condition leads to Clairaut's Law.. a particularly easy relation. Geodesic constant condition  is derivable as Liouville's relation generally for the $k_g=0$ condition from Christoffel symbols.