If I understand the problem correctly, we just need to extract the path from `StreamPlot`, don't we?

    previousCell[] := 
     Cells@EvaluationNotebook[] /. {___, prev_, EvaluationCell[], ___} :> 
       ToExpression@First@NotebookRead@prev
    
    Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.\
    com/eDzXT.png"]

The following part of code should be placed in a new cell under the previous code:
    
    dat = Uncompress@previousCell[];

    func = Interpolation@dat;
    
    {vx, vy} = Function[{x, y}, #] & /@ Grad[func[x, y], {x, y}];
    
    begin = {180.0, 179.99};
    end = {124.5, 124.49};
    
    plot = StreamPlot[{vx[x, y], vy[x, y]}, {x, #, #2}, {y, #3, #4}, StreamPoints -> {begin},
         StreamStyle -> "Line", Epilog -> Point@{begin, end}] & @@ Flatten@func["Domain"]

![Mathematica graphics](https://i.sstatic.net/Ws3Ms.png)
    
    path = Cases[Normal@plot, Line[a_] :> a, Infinity][[1]];
    
    ListPlot3D@dat~Show~Graphics3D@Line[Flatten /@ ({path, func @@@ path}\[Transpose])]

![Mathematica graphics](https://i.sstatic.net/m0wzm.png)