The position vector of a point in space depends on how you define your coordinate system.
Suppose that one coordinate system $S$ defined by $(x,y,z)$ is fixed at the origin. Suppose that a second coordinate system $S_0$ is defined by axes labeled $(x_0,y_0,z_0)$ with origin at the location $\Delta \vec{r}$ relative to the first frame.
A particle is at the location $\vec{r}$ relative to the fixed frame. Relative to the second frame, its location is $\vec{r}_0$. These vectors are related according to
$$\vec{r}=\Delta \vec{r} + \vec{r}_0.$$Suppose that the second reference frame is moving and starts with its origin aligned with the first reference frame at $t=0$. Then the origin of the moving frame $S_0$, relative to the fixed frame, is at the location
$$\Delta \vec{r}= \int \vec{V}d t$$and the position of the particle in the fixed frame is
$$\vec{r}=\vec{r}_0 + \int \vec{V}dt.$$In frame $S$, the particle's velocity is $\vec{v}=\frac{d\vec{r}}{dt}$. In frame $S_0$, the particle's velocity is $\vec{v}=\frac{d\vec{r}_0}{dt}$. So solve the above equation for $\vec{r}_0$ and take the derivative with respect to time.
$$\vec{r}_0=\vec{r} - \int \vec{V}dt$$$$\frac{d\vec{r}_0}{dt}=\frac{d\vec{r}}{dt} - \frac{d}{dt}(\int\vec{V}dt)$$$$\vec{v}_0= \vec{v} - \vec{V}$$We can also calculate the acceleration of the particle measured in the moving frame. Taking the time derivative again gives
$$\frac{d\vec{v}_0}{dt}= \frac{d\vec{v}}{dt} - \frac{d\vec{V}}{dt}$$$$\vec{a}= \vec{a}_0 - \vec{A}$$where $\vec{A}$ is the acceleration of the moving frame (relative to the fixed frame).
If $\vec{V}$ is constant, $\frac{d\vec{V}}{dt}=0$. Then,
$$\frac{d\vec{v}_0}{dt}= \frac{d\vec{v}}{dt}$$$$\vec{a}_0=\vec{a}\, .$$In this case, observers in the two frames will measure the same acceleration for the particle and will agree on Newton's laws (as well as all other laws). Therefore, a reference frame that moves at a constant velocity relative to an inertial reference is also an inertial reference frame.