Finding Point on a Bounds (2D Rectangle) using its Centre and an Angle.

Here i quickly wrote those two helper functions:

 public static Vector2 PointOnBounds(Bounds bounds, Vector2 aDirection)
 {
     aDirection.Normalize();
     var e = bounds.extents;
     var v = aDirection;
     float y = e.x * v.y / v.x;
     if (Mathf.Abs(y) < e.y)
         return new Vector2(e.x, y);
     return new Vector2(e.y * v.x / v.y, e.y);
 }
 public static Vector2 PointOnBounds(Bounds bounds, float aAngle)
 {
     float a = aAngle * Mathf.Deg2Rad;
     return PointOnBounds(bounds, new Vector2(Mathf.Cos(a), Mathf.Sin(a)));
 }

Note: The used Bounds has to be in the same coordinate system as the direction vector / angle. The angle is as always counter clockwise, specified in degree and "0°" points to the right.

Collider.bounds and Renderer.bounds returns the bounds in worldspace. So the box is axis aligned to the world axes. However Mesh.bounds is defined in localspace so the direction vector has to be in localspace as well.

I haven't tested the method. The idea is to project the center along the direction first onto the "x" size of the bounds. If the resulting point is outside the bounds we do the same for the y size. So in your example image the first case would be outside the bounds. If you extend the purple line until it crosses the "x size", the y position would be above the bounds rect (somewhere below your word "step" in the text above). So the second case would apply here.