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"]

path = Cases[Normal@plot, Line[a_] :> a, Infinity][[1]];

ListPlot3D@dat~Show~Graphics3D@Line[Flatten /@ ({path, func @@@ path}\[Transpose])]

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

If you have difficulty in accessing the data in dropbox in the question, try the following:

Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.\
com/eDzXT.png"]
SelectionMove[EvaluationNotebook[], Previous, Cell, 1];
dat = Uncompress@First@First@NotebookRead@EvaluationNotebook[];
NotebookDelete@EvaluationNotebook[];

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"]

path = Cases[Normal@plot, Line[a_] :> a, Infinity][[1]];

ListPlot3D@dat~Show~Graphics3D@Line[Flatten /@ ({path, func @@@ path}\[Transpose])]