A star is just born and starts travelling towards us at 99 percent speed of light. So how would an immortal observer observe the light from it? An immortal observer is observing a new born star from billions of lightyears away travelling towards it at near the speed of light (99%).
The star is a short living one and dies soon before reaching the observer so it doesn't kill him.
How would the observer perceive the light?
Just for a short period of time or longer than the star's life in his reference or shorter time or what?
 A: The star's light would be blueshifted, and if the time it took to go from birth to supernova is $t_{0}$, then it would survive for a period of time equal to
$$t = \frac{t_{0}}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}$$
according to the observer.  If this lengthening of the "decay time" seems weird, note that this has actually been observed for radioactive particles in particle accelerators.  the decay time of, say, a muon, is exactly extended by having the muon move at relativistic speeds relative to the lab reference frame.
A: They will see the star's lifetime as much shorter than its proper lifetime. The frequency of the light from the star will also be multiplied (blueshifted) by the same factor.
At an initial distance of billions of light years, spacetime curvature is significant, and there's no good way to even define what it means for the star to move toward you at $0.99c$. However, if the star arrives at that speed at the end of its trip, then the blueshift factor at the end of the trip will be $\sqrt{\frac{1+v/c}{1-v/c}} \approx 14$.
In the rest of this answer I'll assume special relativity and a shorter initial distance. (Say $D$ is the distance.)
The blueshift will then be $\approx 14$ over the whole trip.
The star's proper lifetime will be shorter than its coordinate lifetime $D/v$ by a factor of $\frac{1}{\sqrt{1-(v/c)^2}} \approx 7$. This stacks with the other factor, so on the whole, the lifetime they see is shorter than $D/v$ by a factor of $100$. (Exactly $100$, since the product of those two ratios is $\frac{1}{1-v/c}$.)
It's easy to see where the factor of $100$ comes from: the star starts moving at $0.99c$ at the same time that the first light from it starts moving at $c$. It takes time $D/c$ for the light to arrive, and $D/v$ for the star to arrive and the light to stop, which is about 1% longer.

The earlier answer suggests that you would see the star live longer than its proper lifetime by a factor of $\approx 7$, which is not correct.
If $D$ is small, then $D/v$ is longer than the proper lifetime by that factor, but that isn't what you see.
If there was a clock moving at the same speed as you at the location where the star appeared, and it was Einstein-synchronized with your wristwatch before the experiment began, and you subtracted the reading on your wristwatch when the star arrived from the reading on the distant clock when the star departed, then you would get $7$ times the proper lifetime of the star, and $100$ times the lifetime you actually saw.
In the cosmological case where $D$ is billions of light years, you couldn't have Einstein-synchronized clocks. You could use cosmological time instead. The coordinate lifetime of the star would still be larger than its proper lifetime, but by a different factor.
A: 
How would the observer perceive the light? Just for a short period of time or longer than the star's life in his reference or shorter time or what?

The relativistic Doppler shift is $$\frac{f_o}{f_s}=\sqrt{\frac{1+v/c}{1-v/c}} \approx 14$$
So the observer would perceive the light blueshifted by a factor of 14. Similarly, the observer would actually see the light for 1/14 of the "normal" lifetime of such stars.
However, the relativistic time dilation factor is $$\gamma = (1-v^2/c^2)^{-1/2} \approx 7$$
So, if the observer were to correct for the light travel delays due to the finite speed of light then they would calculate that the star is actually redshifted by a factor of 7 and actually lives 7 times the "normal" lifetime of such stars.
A: There are two separate factors to consider here, namely the relativistic Doppler effect and time dilation.
Suppose the star's lifetime in its own rest frame is t years.
The direct answer to your question is that owing to the Doppler effect, the observer would see the lifecycle of the approaching star speeded up, so the star would appear to grow and burn itself out in less than t years.
However, in the observers rest frame, the star was born more than t years ago, so the actual lifespan of the star in the observer's frame of reference is more than t years (ie the lifespan is time dilated).
