Covalent Bonding in odd electron sharing molecules

Why in odd electron sharing molecules (like NO ) the bond containing 3 electrons is considired to be half covalent bond?

I will answer to your question, but first let me show you that NO has a $2.5$ bonds, and I will use what I made for NO to answer your question. Here we go ! :)

Proof that NO has a $2.5$ bonds :

First the electron configurations of these two elements are :

$\ce{{}_7 N :1s^2 2s^2 2p^3}$ which have $5$ valence electrons in $4$ orbitals (one $\ce{s}$ and three $\ce{p}$)

$\ce{{}_8 O :1s^2 2s^2 2p^4}$ which have $6$ valence electrons in $4$ orbitals (one $\ce{s}$ and three $\ce{p}$)

Then we can build the orbitals-diagram of $\ce{NO}$ and I will suppose that its not a correlated diagram, and because of electronegativity of these elements verifies $\ce{\chi(N)<\chi(O)}$ then we have:

Then the electon configuration of $\ce{NO}$ is $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)^2 (1 \pi_{x})^2 (1 \pi_{y})^2 (1 \pi_{x}^{\star})}$

The bond order $i$ is : $$i=\frac{\text{number of electrons in bonding MOs}-\text{number of electrons in anti-bonding MOs}}{2}$$

Here we have $$i=\frac{2\times 4-3}{2}=\frac{5}{2}=2.5$$

Quod Erat Demonstrandum.

So now I did it for $\ce{NO}$ I have this configuration : $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)^2 (1 \pi_{x})^2 (1 \pi_{y})^2 (1 \pi_{x}^{\star})}$.

Then you said $3$ electrons bond considered has $0.5$ bond. As I understand your question, its the case if the configuration is $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)}$ (for example but it could be an other) then you have $3$ electrons in bonding bonds and $2$ in anti-bonding bond.

Then the bond order of your molecule is : $$i=\frac{3-2}{2}=0.5$$