6.D.

The length of line segments with grid points as vertices can take 8 values: $1, \sqrt{2}$, $2, \sqrt{5}, 2 \sqrt{2}, 3, \sqrt{10}, \sqrt{13}$. The isosceles right triangles formed by these line segments can be classified into 4 cases based on their side lengths:
$$1,1, \sqrt{2} ; \sqrt{2}, \sqrt{2}, 2 ; 2,2,2 \sqrt{2} ; \sqrt{5}, \sqrt{5}, \sqrt{10} \text {. }$$

Below, we count the triangles by classifying them according to the length of the hypotenuse.
(1) When the hypotenuse is $\sqrt{2}$, the hypotenuse must be the diagonal of a small square. There are 12 such line segments, and each such line segment corresponds to two isosceles right triangles, totaling $2 \times 12=24$.
(2) When the hypotenuse is 2, there are 10 line segments of length 2 in the figure, 6 of which are on the perimeter of a $2 \times 3$ rectangle, each corresponding to one isosceles right triangle; the other 4 are inside the $2 \times 3$ rectangle, each corresponding to two isosceles right triangles. In total, there are $6+2 \times 4=14$.
(3) When the hypotenuse is $2 \sqrt{2}$, the hypotenuse must be the diagonal of a $2 \times 2$ square. There are 4 such line segments, each corresponding to two isosceles right triangles, totaling $2 \times 4=8$.
(4) When the hypotenuse is $\sqrt{10}$, there are 4 such line segments, each corresponding to one isosceles right triangle, totaling 4.

Therefore, the number of isosceles right triangles with grid points as vertices is $24+14+8+4=50$.