Here how we can extract data from NDSolveValue

exact[x_, y_] := Sin[\[Pi] x] Sin[\[Pi] y]
exactData = Partition[{{#1, #2}, exact[#1, #2]} & @@@ pts // Flatten, 3] // N
numData = Partition[NDSolveValue[{op ==0., \[CapitalGamma]}, {{#1, #2}, u[#1, #2]} & @@@ pts, {x, y} \[Element] mesh]//Flaten, 3] // N
ListPointPlot3D[{exactData,  numData }]

Yes, order of vertex coordinate of each triangle must be in counterclockwise. Thanks for your helps. Here is my final result. Now how can I extract nSol result for some values from NDSolveValue.? I mean instead of graphing it I need data to find error. By the way is there a good way to use Module function to calculate nSol for different n values. n=8,16,32,64. Thanks.

Needs["NDSolve`FEM`"]

n = 8;
pts = Join @@ Table[{i, j}, {j, 0, 1, 1/n}, {i, 0, 1, 1/n}];
upperLeft =1/n Flatten[Drop[Table[{{i, j}, {i + 1, j + 1}, {i, j + 1}}, {j, 0, n}, {i, 0,n}], -1, -1], 1];
lowerRight = 1/n Flatten[Drop[Table[{{i, j}, {i + 1, j}, {i + 1, j + 1}}, {j, 0, n}, {i, 0,n}], -1, -1], 1];
triangs = Riffle[upperLeft, lowerRight];
indxRules = MapIndexed[Rule[#1, First[#2]] &, pts];
element = triangs /. indxRules;
mesh = ToElementMesh["Coordinates" -> pts, "MeshElements" -> {TriangleElement[element]}] // Quiet;
mesh["Wireframe"];

op = -Laplacian[u[x, y], {x, y}] - 2 \[Pi]^2 Sin[\[Pi] x] Sin[\[Pi] y];
\[CapitalGamma] = DirichletCondition[u[x, y] == 0., True];
nSol = NDSolveValue[{op == 0., \[CapitalGamma]}, u[x, y], {x, y} \[Element] mesh];

Plot3D[nSol, {x, 0, 1}, {y, 0, 1}]

ContourPlot[nSol, {x, y} \[Element] mesh, AspectRatio -> Automatic]

Show[ContourPlot[nSol, {x, y} \[Element] mesh,  AspectRatio -> Automatic], mesh["Wireframe"]]

We have a problem here. My ultimate goal is to solve Poisson equation $$-\Delta u=f,\quad(x,y)\in\Omega=(0,1)\times(0,1)$$ $$u=0,\quad \Gamma=\partial\Omega$$ where $f(x,y)=2\pi^2\sin(\pi x)\sin(\pi y)$ using FEM and grid as above. (Exact solution is $u(x,y)=\sin(\pi x)\sin(\pi y )$)

Let's look at Mathematica's example..

op = -Laplacian[u[x, y], {x, y}] - 20
\[CapitalGamma] = DirichletCondition[u[x, y] == 0, x == 0 || x == 2]
 mesh = ToElementMesh["Coordinates" -> {{0., 0.}, {1., 0.}, {2., 0.}, {2., 1.}, {1.,1.}, {0., 1.}},   "MeshElements" -> {TriangleElement[{{1, 2, 5}, {5, 6, 1}, {2, 3,4}, {4, 5, 2}}]}]
ufun = NDSolveValue[{op == 0, \[CapitalGamma]},  u[x, y], {x, y} \[Element] mesh];
Show[ ContourPlot[ufun, {x, y} \[Element] mesh, AspectRatio -> Automatic],
 mesh["Wireframe"]]

it solves the problem..

But when I make the same mesh using below code, it is not working

n = 2;(*to be choosen*)
pts = Flatten[Table[{x, y}, {y, 0, n - 2}, {x, 0, n - 1}], 1];
indexPts = Flatten[Table[{x, y}, {y, 0, n - 1}, {x, 0, n}], 1];
mkTriangs[pt : {x_, y_}] := 
 Sequence[{pt, pt + {0, 1}, pt + {1, 1}}, {pt, pt + {1, 0}, 
   pt + {1, 1}}]
triangs = mkTriangs /@ pts;
indxRules = MapIndexed[Rule[#1, First[#2]] &, indexPts];
index = triangs /. indxRules;
index = triangs /. indxRules;
Needs["NDSolve`FEM`"]
mesh = ToElementMesh["Coordinates" -> indexPts, 
    "MeshElements" -> {TriangleElement[index]}] // Quiet;
mesh["Wireframe"]

Mathematica uses outside first, I don't know it is matter. If it is matter how can I get this order..

ToElementMesh["Coordinates" -> {{0., 0.}, {1., 0.}, {2., 0.}, {2., 1.}, {1.,1.}, {0., 1.}},   "MeshElements" -> {TriangleElement[{{1, 2, 5}, {5, 6, 1}, {2, 3,4}, {4, 5, 2}}]}]

Any suggestion..

i didn't see your reply, before I posted mine, sorry about that. Very similar code!

Hi, here is a very straight forward approach:

n = 10;  (* to be choosen *)
pts = Flatten[Table[{x, y}, {y, 0, n}, {x, 0, n}], 1];
indexPts = Flatten[Table[{x, y}, {y, 0, n + 1}, {x, 0, n + 1}], 1];

mkTriangs[pt : {x_, y_}] := 
 Sequence[{pt, pt + {0, 1}, pt + {1, 1}}, {pt, pt + {1, 0}, pt + {1, 1}}]

triangs = mkTriangs /@ pts;
indxRules = MapIndexed[Rule[#1, First[#2]] &, indexPts];
index = triangs /. indxRules;

Needs["NDSolve`FEM`"]
elementMesh = 
  ToElementMesh["Coordinates" -> indexPts, "MeshElements" -> {TriangleElement[index]}] // Quiet;

elementMesh["Wireframe"]

which results in:

Regards -- Henrik