What kind of a visual trajectory would an asteroid have if it were to hit Earth? Let's say that Peter has a set of superb telescopes and he's watching the sky every night, for 3-4 months. He may not know it just yet, but there is an asteroid heading towards Earth, on a crash course. He is too lazy to plot an orbit, because he doesn't believe he could die like that.

Over the months the Earth travels quite a lot around the Sun, resulting in various amounts of parallax, depending on the distance. The location of the telescopes are also a factor, and so is the (rotation of the) axis of Earth.

To simplify this and to get rid of Earth radius parallax, Peter is a researcher on the South Pole with those mighty telescopes.
Over the course of months, what kind of a trajectory would a doomsday asteroid make? As it gets bigger/brighter (which is also relative, because the asteroid itself can rotate)/closer, how would its position change, compared to the stars in the background?
Would the change slow down or accelerate? Would it be a small piece of a spiral (I don't mean complete spirals, just the fact that it wouldn't be a parabola or a circle)? What do you think Peter would see if he animated its movement during the course of months on a virtual sky sphere?

The National Geographic article Asteroid Called ‘Spooky’ Will Buzz Earth on Halloween has an animation of the trajectory of asteroid 2015 TB145 (nicknamed Spooky because it looks like a skull).

16 minutes apart visual timelapse:
https://upload.wikimedia.org/wikipedia/commons/e/ec/2015_TB145_discovery.gif
There is no simple answer to this question other than: you have to do the math.

Most often you would see the asteroid brightening more rapidly and moving across the sky faster as the days pass, but not necessarily.  It could arrive directly along our orbital path and just brighten but show no motion in the sky.  It could start out lit up fully (full phase) and then arrive with only partial lighting from the sun, so it does not necessarily brighten.   

You generally would not know the phase, unless you calculate it, because you will not be able to resolve it until it gets really close.