Explaination on phasor diagram for RL circuit?
shouldn't by theory, when inductance is increase, thus the degree for
Vl also gets bigger?

For Vl read \$V_L\$.
No, think about what happens when \$R\$ gets smaller and smaller until it eventually equals zero ohms. At that point, the voltage across the inductor (\$V_L\$) equals \$V_S\$ i.e. the angle gets smaller until it equals 0° relative to \$V_S\$.
This can also be seen in the angle of the current. As \$R\$ gets smaller, the current becomes lagging more and more and, when \$R\$ is zero ohms, the current in the series circuit is lagging by 90° and, of course this means that the voltage is leading current by 90° and is at 0° relative to the supply voltage.
If you'd used better quality numbers or better quality calculations you'd find that the two angles shown in each diagram add-up to exactly 90°. Your first diagram appears to be in error in that respect.