You are correct, her mass did not change. And while gravity is still acting on her, what she did was use her legs muscles to store energy (like springs) such that the resulting spring force was greater than the force caused by her mass (weight).
So if you look at the force balance you get:
$$\text{Leg Force} - \text{Weight} = \text{mass} \cdot \text{acceleration}$$
Where $\text{Weight} = \text{mass} \cdot a_{\text{gravity}} $
So when $\text{Leg Force} > \text{Weight}$ you get a net positive acceleration in the direction opposite gravity (up) which causes her to "jump" off your back.
A weighing machine measures the force exerted by a body on the weighing machine.
Newton's third law then predicts that there is a force of the same magnitude and opposite in direction acting on the body producing the force.
On the Earth if the weighing machine and the body are not accelerating (ignoring the rotation of the Earth) then the reading on the weighing machine will be the weight of the body.
If the weighing machine and the body are accelerating then you could call the reading on the weighing machine the apparent weight of the body.
So including the effect of the rotation of the Earth it is only at the geographic poles that reading on the weighing machine is the weight of the body.
Elsewhere on the Earth the reading on the weighing machine will be lower than at the poles so you could call that the apparent weight.
The difference between these readings is small.
If the weight of the body is $10 \, \rm N$ then with the weighing machine and the body in a stationary lift, or a lift moving at constant velocity upwards or downwards the reading on the weighing machine would be $10 \, \rm N$ which is the weight of the body.
If the weighting machine and the lift had an upward acceleration of $5 \,\rm m s^{-2}$ then the reading on the weighing machine would be $15 \, \rm N$ and you could say that the apparent weight of the body was $15 \, \rm N$
If the weighting machine and the lift had a downward acceleration of $5 \,\rm m s^{-2}$ then the reading on the weighing machine would be $5 \, \rm N$ and you could say that the apparent weight of the body was $5 \, \rm N$
If the weighting machine and the lift had a downward acceleration of $10 \,\rm m s^{-2}$ then the reading on the weighing machine would be $0 \, \rm N$ and you could say that the apparent weight of the body was zero - the body appeared to be weightless.
The definition of weight that I have used is that the weight of a body is the force on the body due to the gravitational attraction of the Earth.
However others define the weight of a body as the reading on a weighing machine as explained by Walter Lewin in one of his 8.01 Classical Mechanics lectures.
Using this definition a body is weightless when it is in free fall.
Best Answer
Suppose you throw the ball upwards at some speed $v$. Then the time it spends in the air is simply:
$$ t_{\text{air}} = 2 \frac{v}{g} $$
where $g$ is the acceleration due to gravity. When you catch the ball you have it in your hand for a time $t_{\text{hand}}$ and during this time you have to apply enough acceleration to it to slow the ball from it's descent velocity of $v$ downwards and throw it back up with a velocity $v$ upwards:
$$ t_{\text{hand}} = 2 \frac{v}{a - g} $$
Note that I've written the acceleration as $a - g$ because you have to apply at least an acceleration of $g$ to stop the ball accelerating downwards. The acceleration $a$ you have to apply is $g$ plus the extra acceleration to accelerate the ball upwards.
You want the time in the hand to be as long as possible so you can use as little acceleration as possible. However $t_{\text{hand}}$ can't be greater than $t_{\text{air}}$ otherwise there would be some time during which you were holding both balls. If you want to make sure you are only ever holding one ball at a time the best you can do is make $t_{\text{hand}}$ = $t_{\text{air}}$. If we substitute the expressions for $t_{\text{hand}}$ and $t_{\text{air}}$ from above and set them equal we get:
$$ 2 \frac{v}{g} = 2 \frac{v}{a - g} $$
which simplifies to:
$$ a = 2g $$
So while you are holding one 3kg ball you are applying an acceleration of $2g$ to it, and therefore the force you're applying to the ball is $2 \times 3 = 6$ kg.
In other words the force on the bridge when you're juggling the two balls (with the minimum possible force) is exactly the same as if you just walked across the bridge holding the two balls, and you're likely to get wet!