Visualizing linear transformations on vector fields

To visualize the transformation of a vector field you first have to visualize the vector field. In your case, you are rotating the identity field in $\mathbb R^2$.

Consider an arbitrary point in $\mathbb R^2$, say $(3,4)$. The identity vector field assigns the vector $3\hat i + 4\hat j$ to this point. You imagine an arrow of length $5$ being drawn with its tail at the point $(3,4)$. You perform a similar construction for each point in the plane. Since you cannot do that in a finite period of time you plot enough of these in your mind or on paper until you get a satisfactory picture of the vector field.

To visualize what happens under the transformation, you apply the transformation to the vectors you have drawn. You replace the vectors that were previously drawn with the new vectors you get from the transformation. That will give you an idea of what the new vector field will look like.

If you try this exercise by hand you will discover why the $60^{\circ}$ rotation looks different.

(I'll keep it simple, so this explanation sacrifices a lot of generality.)

There are two ways to apply a linear transformation to a vector field: multiplication and conjugation. What you've observed is the difference between the two.

Multiplying a vector field (on the left) rotates all the arrows in-place. If you start with a vector field that looks like circles, and you rotate all the arrows a bit, they no longer go in circles; they either spiral into or out of the origin. This is the relationship between your two versions of $T$; you can write $T_2 = RT_1$ for some rotation $R$.

It sounds like you expected the vector field to look the same, just rotated in space. That would indeed be the case if you had conjugated $T$ by a rotation, getting $T_3=RT_1R^{-1}$. In fact, since 2D rotations commute with each other, you would have $T_3=T_1$: the two vector fields would look exactly the same.