So the spacetime velocity refers to the 4-velocity. Most of our favorite 3D vectors have a 4-vector counter-part, and it is this 4-vector counter part Lorentz transforms between various interial frames.
The most basic one is spatial separation:
$$ \vec r = (\Delta x,\Delta y, \Delta z)$$
Different coordinate systems have different values of the $\Delta$'s, but they all agree on the length:
$$||\vec r||^2 = \Delta x^2+\Delta y^2+ \Delta z^2$$
in 3D space. The 4-vector version is:
$$ r^{\mu} = (c\Delta t, \vec r)$$
All inertial frames agree on the so-called invariant interval:
$$\Delta s^2 = r^{\mu}r_{\mu} \equiv c^2\Delta t^2 - ||\vec r||^2 $$
If you differentiate a world-line defined by $r^{\mu}(\tau)$ with respect to $\tau$, you get the 4-velocity:
$$ \frac{dr^{\mu}}{d\tau} = v^{\mu} = (\gamma c, \vec v) $$
where $\gamma=1/\sqrt{1-(/c)^2}$ is the usual Lrentz factor. It's magnitude is:
$$v^{\mu}v_{\mu} = \gamma^2c^2 - \gamma||\vec v||^2 = c^2$$
which is constant for all $\vec v$. This is the origin of "we all move through spacetime at the speed of light".
In the rest frame, it is:
$$ v^{\mu} = (c, 0, 0,0) $$
That is, "we move through time at $c$".
Now there is no frame that moves at $c$, and we can't write down the 4-velocity of light because $\gamma \rightarrow \infty$, but as you approach $c$, you don't move through space only (your point 2).
We can rewrite it (in the $z$-direction) as
$$ v^{\mu} = \gamma c(1, 0, 0, \sqrt{1-1/{\gamma^2}}) \rightarrow \gamma c (1, 0, 0, 1) $$
so light moves through space and time equally. The scale factor $c\gamma$ diverges, which is why people say light doesn't move through time, and sees the Universe Lorentz contracted into a 2D plane, but that view point gives no insight into relativity. It is a viewpoint from an inertial frame that does not exist.
As seen in the twin paradox, the path between two time-like ($\Delta s^2 > 0$) events is the inertial path in-which both events occur at the same location. That way $\Delta r=0$, so that:
$$ \Delta s^2 = \Delta t^2 $$
and $\delta t^2$ is maximized. (Remember, all inertial frames see the same $ \Delta s^2$).
