You just need to apply the Angle Bisector Theorem again. For completeness, I'll go through the whole argument.
Given $\triangle ABC$ with side-lengths $a = |\overrightarrow{BC}|$, $b = |\overrightarrow{CA}|$, $c := |\overrightarrow{AB}|$, suppose the angle bisector from $A$ meets the opposite side at $D$. You're correct that, by the Angle Bisector Theorem
$$\frac{|\overrightarrow{BD}|}{|\overrightarrow{DC}|} = \frac{|\overrightarrow{AB}|}{|\overrightarrow{AC}|} = \frac{c}{b} \tag{$\star$}$$
so that we can write:
$$D = \frac{B b + C c}{b+c}$$
Note also that $(\star)$ implies
$$\frac{|\overrightarrow{BD}|}{|\overrightarrow{BC}|} = \frac{c}{b+c} \qquad\text{so that}\qquad |\overrightarrow{BD}| = |\overrightarrow{BC}|\frac{c}{b+c} = \frac{ac}{b+c}$$
Now, here's your next step: Suppose the angle bisector from $B$ meets $\overline{AD}$ at $X$ (which we "know" is the incenter). Again by the Angle Bisector Theorem applied to $\triangle ABD$,
$$\frac{|\overrightarrow{AX}|}{|\overrightarrow{XD}|} = \frac{|\overrightarrow{BA}|}{|\overrightarrow{BD}|} = \frac{c}{ac/(b+c)} = \frac{b+c}{a}$$
and we can write
$$X = \frac{A a + D (b+c)}{a+b+c} = \frac{A a + B b + C c}{a+b+c}$$
Because this formula is symmetric in the elements of $\triangle ABC$, we conclude that $X$ is not merely where the angle bisector from $A$ meets the angle bisector from $B$, but where any two angle bisectors meet ... and, therefore, where all three bisectors meet (aka, the incenter). $\square$
Best Answer
It is important that you first normalize vectors $\vec{OA} \, ( = \vec{a})$ and $\vec{OB} \, ( = \vec{b})$. Please see the below diagram as an example. The directional vector of angle bisector is in direction $OE$ but if you do not normalize the vectors, you will get a vector in direction $OF$.
So, $\vec{OE} = \vec{OC} + \vec{OD}$
Unit vector $\vec{OC} = \displaystyle \frac{\vec{a}}{||a||}$
Unit vector $\vec{OD} = \displaystyle \frac{\vec{b}}{||b||}$
So, $\vec{OE} = \displaystyle \frac{\vec{a}}{||a||} + \frac{\vec{b}}{||b||}$
Directional vector for one of the angle bisectors = $\lambda_1 (||b|| \, \vec{a} + ||a|| \, \vec{b})$
Similarly the directional vector of the other angle bisector = $ \lambda_2 (||a|| \, \vec{b} - ||b|| \, \vec{a})$
Dot product of both angle bisectors = $\lambda_1 \lambda_2 \, (||b|| \, \vec{a} + ||a|| \, \vec{b}) \cdot (||a|| \, \vec{b} - ||b|| \, \vec{a}) = 0$
(as $\vec{a} \cdot \vec{a} = ||a||^2 \,$ and $ \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$)