Given points

$O=(0,0), \quad A=(1,2) \Rightarrow z_1=1+2i$

Let $B=(x,y) \Rightarrow z_2=x+iy, \quad x<0$

We are told that $\triangle OAB$ is a right-angled isosceles triangle with $OB$ as hypotenuse.

Interpret the geometry

Since $OB$ is the hypotenuse, the right angle is at $A$.

Also, because the triangle is isosceles, the two equal sides are: $OA = AB$

Now, $OA = |z_1| = \sqrt{1^2+2^2}=\sqrt{5}$

So we must have $AB=\sqrt{5}$ and $\angle OAB=90^\circ$

Use vector form

At point $A$, the vectors are: $\overrightarrow{AO}=(-1,-2), \quad \overrightarrow{AB}=(x-1,y-2)$

Since they are perpendicular, $(-1,-2)\cdot(x-1,y-2)=0$ $-(x-1)-2(y-2)=0$ $-x+1-2y+4=0$ $x+2y=5 \quad ...(1)$

Also, $AB=\sqrt5$, so $(x-1)^2+(y-2)^2=5 \quad ...(2)$

Solve for $B$

From (1): $x=5-2y$

Put into (2): $((5-2y)-1)^2+(y-2)^2=5$ $(4-2y)^2+(y-2)^2=5$ $4(2-y)^2+(y-2)^2=5$ $5(y-2)^2=5$ $(y-2)^2=1$ $y=3 \text{ or } 1$

If $y=3$, then $x=5-2(3)=-1$ so $z_2=-1+3i$

If $y=1$, then $x=5-2(1)=3$ so $z_2=3+i$

But given $\operatorname{Re}(z_2)<0$, we take $z_2=-1+3i$

Now check each option

Option A

$z_2=-1+3i$ This lies in quadrant II, so $\arg z_2 = \pi - \tan^{-1}\left(\frac{3}{1}\right)=\pi-\tan^{-1}3$ So A is true.

Option B

Compute: $z_1-2z_2=(1+2i)-2(-1+3i)=1+2i+2-6i=3-4i$ Hence $\arg(z_1-2z_2)=\arg(3-4i)=-\tan^{-1}\left(\frac{4}{3}\right)$ So B is true.

Option C

$|z_2|=|-1+3i|=\sqrt{(-1)^2+3^2}=\sqrt{10}$ So C is true.

Option D

Compute: $2z_1-z_2=2(1+2i)-(-1+3i)=2+4i+1-3i=3+i$ Therefore, $|2z_1-z_2|=|3+i|=\sqrt{3^2+1^2}=\sqrt{10}$ which is not equal to $5$.

So D is NOT true.

Final conclusion

The statement which is NOT true is: $\boxed{\text{D}}$