There are excellent answers already, but I wanted to address the following question

[...] how would we mathematically define a normal coordinate for the oxygen(s) and hydrogen during a normal mode vibration of water molecule?

The first step is to switch from cartesian coordinates to internal coordinates. For water, it would be the two bond lengths $L_1$ and $L_2$ and the bond angle $\alpha$. This separates out translations and rotations (internal coordinates are not affected by those degrees of freedom). The second step is to turn these internal coordinates into normal coordinates so that the conformational energy is minimal when all normal coordinates are zero, and so that each normal coordinate corresponds to a normal mode.

I will call the normal coordinates $q_s$, $q_a$ and $q_b$, for symmetric stretch, asymmetric stretch and bend (symmetric), respectively. Separately, they are scalars. Also, let's call the lowest energy bond angle $\alpha_0$ and the lowest energy bond length $L_0$.

For small amplitudes, the bond lengths don't change for the symmetric bend. We can simply define $q_b$ as:

$$q_b = \alpha - \alpha_0$$

For the symmetric stretch, the sum of the bond lengths changes, so we can define:

$$q_s = L_1 + L_2 - 2 L_0$$

For the asymmetric stretch, the difference of the bond lengths is nonzero, so we can define:

$$q_a = L_1 - L_2$$

This should work for small amplitudes, again. Now, you can express the conformational energy of the water molecule using expressions containing the square of the normal coordinates, separately for each normal mode.

Conversely, we can explore how bond angles and lengths change depending on the values of the normal coordinates:

$$\alpha = q_b + \alpha_0$$

$$ L_1 = \frac{q_s + q_a}{2} + L_0$$

$$ L_1 = \frac{q_s - q_a}{2} + L_0$$

So you can see that as a first approximation, the normal coordinates are linear combinations of the internal coordinates and vice versa. It gets more complicated if there are torsion angles and tetrahedral or octahedral centers.

Water is a good example for figuring out the logic of normal coordinates on paper. It is neither too simple nor too complex.

There are excellent answers already, but I wanted to address the following question

[...] how would we mathematically define a normal coordinate for the oxygen(s) and hydrogen during a normal mode vibration of water molecule?

The first step is to switch from cartesian coordinates to internal coordinates. For water, it would be the two bond lengths $L_1$ and $L_2$ and the bond angle $\alpha$. This separates out translations and rotations (internal coordinates are not affected by those degrees of freedom). The second step is to turn these internal coordinates into normal coordinates so that the conformational energy is minimal when all normal coordinates are zero, and so that each normal coordinate corresponds to a normal mode.

I will call the normal coordinates $q_s$, $q_a$ and $q_b$, for symmetric stretch, asymmetric stretch and bend (symmetric), respectively. Separately, they are scalars. Also, let's call the lowest energy bond angle $\alpha_0$ and the lowest energy bond length $L_0$.

For small amplitudes, the bond lengths don't change for the symmetric bend. We can simply define $q_b$ as:

$$q_b = \alpha - \alpha_0$$

For the symmetric stretch, the sum of the bond lengths changes, so we can define:

$$q_s = L_1 + L_2 - 2 L_0$$

For the asymmetric stretch, the difference of the bond lengths is nonzero, so we can define:

$$q_a = L_1 - L_2$$

This should work for small amplitudes, again. Now, you can express the conformational energy of the water molecule using expressions containing the square of the normal coordinates, separately for each normal mode.

Conversely, we can explore how bond angles and lengths change depending on the values of the normal coordinates:

$$\alpha = q_b + \alpha_0$$

$$ L_1 = \frac{q_s + q_a}{2} + L_0$$

$$ L_2 = \frac{q_s - q_a}{2} + L_0$$

So you can see that as a first approximation, the normal coordinates are linear combinations of the internal coordinates and vice versa. It gets more complicated if there are torsion angles and tetrahedral or octahedral centers.

Water is a good example for figuring out the logic of normal coordinates on paper. It is neither too simple nor too complex.