sph=ParametricPlot3D[{2 Cos[u] Sin[v],2 Sin[u] Sin[v], 2 Cos[v]},{u,0,2 Pi},{v,0,Pi}];
np={0,0,2};
tv=vectorPlot[{np,np,np,np},{np+{2,2,0},np+{2,-2,0},np+{-2,2,0},np+{-2,-2,0}}];
Show[tv,sph];
This should illustrate, however, that there is an unique tangent PLANE at each point on the surface. How do we find it? Recall [ vectors.ma notebook: (p-p0).n=0 ] that to find the equation of a plane you need a point on the plane and a normal vector to the plane at that point.
Let's use the north pole of the sphere as our point. Clearly a tangent vector at that point would be (0,0,1). The tangent plane is therefore ( p-(0,0,2) ).(0,0,1)=0 or z=2
tp0=Plot3D[2,{x,-2,2},{y,-2,2}];
Show[sph,tp0];
Show[sph,tp0,ViewPoint->{2,2,0}];Up to Tangent Planes