I thought that the mass of the bike and the gravitational pull would remain constant.
They do. However, the center of mass (or center of gravity) is no longer directly above and in line with the wheels, so part of its effect is "diverted" from pressing the tires straight down. In the crude illustration below:
* = center of mass / center of gravity of bike and rider
[X] is the tire
| is the line of force / weight. The length of the line is proportional to the force.
When riding upright, everything lines up:
*
|
|
|
|
V
[X]
When leaning in a turn, the c/g or c/m moves off the vertical line that passes through the contact patch. Since the distance between the center of gravity and the contact patch is fixed it forms the hypotenuse of the triangle implied in the diagram below (the line directly connecting * and [X]).
*
|
|
|
V<--[X]
Since the force of gravity always runs between the center of the earth and the center of gravity of an object, moving the center of gravity such that the line no longer passes through the tire reduces the amount of the weight that is applied to the tire, by the difference in length of the vertical lines in the two diagrams.
Or so I believe I learned in my physics classes, too many decades ago to admit.
I like a rather large safety margin for stuff that I can't see ahead
I'm right with you on that point. I'm about the slowest rider through curves and corners. I have a morbid fear of a washout and low-side, and even worse of a high-side get-off.