Let's measure our vectors in meters, just to pick a convention (afterwards I'll do it in seconds so that you can see that it doesn't affect the basic ideas).
Given a point in spacetime $(ct_1,x_1,y_1,z_1)$ and another point $(ct_2,x_2,y_2,z_2)$, we can easily construct three things.
1) A vector $\vec{v}=(c(t_2-t_1),x_2-x_1,y_2-y_1,z_2-z_1)$ that points from one to the other.
2) The proper length $l=\sqrt{c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2}$ of that timelike vector $\vec{v}$.
3) And finally, the unit vector $\vec{u}=\vec{v}/l$, which equals $\frac{(c(t_2-t_1),x_2-x_1,y_2-y_1,z_2-z_1)}{\sqrt{c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2}}$.
As numbers (scalars) and vectors (geometric displacements), everyone will agree. The vectors might appear as different quadruples of numbers to different observers (imagine picking a different direction to call your $x$-axis, the geometric entity is the same displacement, but you get different numbers).
Now if you didn't want to measure in meters you can do it again, dividing everything by $c$, except it turns out $u$ will be the same.
1) A vector $\vec{V}=(t_2-t_1,(x_2-x_1)/c,(y_2-y_1)/c,(z_2-z_1)/c)$ that points from one to the other.
2) The proper length $L=\sqrt{(t_2-t_1)^2-(x_2-x_1)^2/c^2-(y_2-y_1)^2/c^2-(z_2-z_1)^2/c^2}$ of that timelike vector $\vec{V}$.
3) And finally, the unit vector $\vec{U}=\vec{V}/L=\vec{u}$, which again equals $\frac{(c(t_2-t_1),x_2-x_1,y_2-y_1,z_2-z_1)}{\sqrt{c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2}}$.
So you have a vector $\vec{v}$ between two spacetime points, as a vector it has a length $l$, and a corresponding unit vector $\vec{u}=\vec{v}/l$ that keeps track of the direction in spacetime, but ignores how much dispalcement (or even which units you use, just the direction in spacetime).
OK, so now lets get to your question. Given two points we have the spacetime displacement $\vec{v}$, the unit displacement $\vec{u}$ that tells us which direction in spacetime to go (but has no informative length or even units), and the length $l$ telling us how far in spacetime to go (and has units of length or time). When someone says proper time, they are referring to the length of the timelike vector, maybe in units of time instead of length.
A common reason people say proper length for spacelike entities and proper time for timelike entities in textbooks isn't because they care a great deal about the units of the answer (it's just a measure of length) but because the squared length $l^2=c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2$ is what is easy to compute and for a timelike spacetime displacement $c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2$ is positive, whereas for a spacelike displacement $-c^2(t_2-t_1)^2+(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2$ is positive. We like to have square roots of positive numbers, so people like to have formulas like $\tau=c^-1\sqrt{c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2}$ and $l=\sqrt{-c^2(t_2-t_1)^2+(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. The only real reason to have two formulas is because then you are taking the square root of a positive number. The idea for either case is the same: the displacement has a magnitude (proper length or proper time) and a direction (unit vector in spacetime), their product gives you the displacement.
Just like in three-space, a vector $(\Delta x,\Delta y,\Delta z)$, has a magnitude, $r=\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}$, and a unit vector (for direction) $((\Delta x)/r,(\Delta y)/r,(\Delta z)/r)$.
Edit: I guess I didn't actually answer the questions.
First question: Is there something missing for proper time being a vector?
Yes, proper time is a magnitude, not a vector, it is just a length in spacetime, it by itself has no direction, there are many directions you could travel all with the same length/magnitude.
Second question: What else is this vector pointing in time direction with the magnitude of proper time?
The vector pointing from one event to the other will look like $(\tau,0,0,0)$ in the frame that inertially moves from one event to the other. But the vector $(\tau,0,0,0)$ is different than the scalar $\tau$. It's like if you walked north a distance of five meters, the distance is 5, everybody agrees. And to you the space vector might be $(5,0,0)$ but what if someone else like to put east first and north second? Then they will say that the vector is $(0,5,0)$. A vector has to tell you the direction. If all you tell me is you walked 5 units I can't tell if you walked $(5,0,0)$, $(0,5,0)$, $(0,0,5)$, or even $(3,4,0)$. The vector tells you the direction and the magnitute, but only when you say what the 3 (or 4) numbers mean first.
