Does it take light longer to reach me if I'm moving away? I'm having trouble grasping the intuitions behind the second postulate of special relativity, particularly what it implies.

*

*For example, imagine that a laser pointer is aimed at me at 1 lightsecond away. Then, I move away from it at a speed of 0.5c. Does the laser still take 1 second to reach me?

I presume the answer is yes. In that case, consider the following case:


*A moves at 0.5c and B is stationary. A shoots a laser from its frame of reference where it is at rest, and measures the time it took the light to travel 1 lightsecond in front of A. This time is also measured by B, but in his frame of reference the light travels a longer distance because A is also moving (thus to be ''1 lightsecond in front of A''  requires light to travel more than 1 lightsecond). Would they measure the same time?

If the answer to question 1 is true, that would mean the answer to question 2 is also true (correct me if this is a non-sequitur), which violates the second postulate of special relativity.
I think I have some gaps in my understanding.
 A: 
For example, imagine that a laser pointer is aimed at me at 1 lightsecond away. Then, I move away from it at a speed of 0.5c. Does the laser still take 1 second to reach me?

It takes 2 seconds according your buddy that you leave behind standing still. Because the buddy sees the distance shrinking at rate 0.5 c.
According to said buddy during those 2 seconds your clock has proceeded 2 seconds multiplied by time dilation factor, so it took 1.732 seconds of time according your clock. You will agree that it took 1.732 second.
As some extra information let me tell that the distance between the point were the beam was emitted and the point were the beam was absorbed is 1.732 lightseconds according to you.
A: The answer to your first question is that in your frame of reference the light will take one second to reach you. The answer to the second is that B will measure a longer time than you did, because light has had further to travel in B's frame.
Those effects arise from the relativity of simultaneity. If you are moving relative to B, then a plane of constant time in your frame will be a sloping slice through time in B's frame, and vice versa. All of the main effects of relativity, such as time dilation and length contraction, arise from that.
Suppose you are standing next to B who flashes a light at t=0 that heads off to the left and right to meet two detectors a light second away. `In B's frame, the light will hit the left and right detector simultaneously after 1s. If  you are travelling towards the right hand detector when B flashes the light, you will see the light hit the right hand detector slightly before it hits the left hand detector. So when it is t=1s at both detectors in B's frame, in your frame it is slightly before t'=1s at the right hand detector and slightly after t'=1s at the left hand detector.
Time in your frame is systematically out of synch with time in B's frame, the size of the effect increasing with distance. In the direction in which you are headed, clocks in B's frame seem progressively ahead of clocks in your frame, while in the opposite direction the clocks in B's frame seem progressively behind clocks in your frame. If you think about that for long enough you should be able to work out that it explains how time dilation can be entirely symmetrical between you and B. As you travel through B's frame you see the clocks you pass getting progressively further ahead than your watch, which makes it appear that your watch is running slow. Meanwhile B, travelling in the opposite direction in your frame, sees each of your clocks getting further and further ahead, giving the impression that B's watch is running slow. In fact, all of your clocks and watches are ticking at the same rate, but all out of synch.
A: 
Does the laser still take one second to reach me?

"One second" as measured by whom? This is the key question. Time is not measured to be the same by all observers; or more accurately, time and space are split up differently by observers moving relative to one another. If I consider myself "stationary", then the ticks of the clock I'm holding take up only time, and no space. A clock that's moving relative to me, though, takes up some space in its ticks (at the beginning of the tick it's in one position, and the end of the tick it's in a different position). That means its time axis is pointing differently through spacetime than my clock's time axis.
So, generally speaking, different observers assign different time and space coordinates to different events. They all agree about one thing (the speed of a beam of light) but differ on how far the light traveled and how much time it took to get there.
A: The second postulate says that the speed of light is constant with respect to any inertial frame, not with respect to you, or any other localized object.
Inertial frames are not local things. They're infinite networks of clocks and metersticks. They never accelerate; they have a constant velocity for all time.
All inertial frames are equivalent, which means you can use any inertial frames you like when solving any problem and you'll get the same answer. You are not obligated to use a frame in which you or any other object is at rest at any particular time.

a laser pointer is aimed at me at 1 lightsecond away.

You need to specify a reference frame with respect to which this distance is measured. You probably mean that you and the laser pointer are at relative rest before you accelerate, and the distance is wrt your mutual rest frame.

Then, I move away from it at a speed of 0.5c.

The answer to the problem of course depends on when you accelerate. If your idea is that you accelerate at the same time the laser pointer is turned toward you, then you have to specify a reference frame wrt which it's the same time. Probably you intend it to be that same pre-acceleration rest frame.
Wrt that frame, the relative speed of you and the light after you accelerate is $c-c/2=c/2$ (yes, it's correct to just subtract), so the light takes 2 seconds of coordinate time to reach you. To get the proper (wristwatch) time you need to divide by $γ=2/\sqrt3$; the result is $\sqrt3$ seconds.
You can also solve this using your post-acceleration rest frame. I'll take $x=t=0$ in that frame to be the place and time where you accelerate. Wrt those coordinates, the laser pointer is turned toward you at $x=2/\sqrt3$ light seconds and $t=1/\sqrt3$ seconds (calculated using the Lorentz transformation). After accelerating, you're stationary at the origin, and the light reaches you at coordinate time $t+x/c=\sqrt3$, which is also your wristwatch time.
With more effort, you can solve the problem using any other inertial frame, and you'll get the same wristwatch time.

A shoots a laser from its frame of reference [...] This time is also measured by B, but in his frame of reference [...]

There's no such thing as the laser's frame of reference or B's frame of reference. Any reference frame is the frame of reference of anyone who chooses to use it as a frame of reference. Coordinate distances and coordinate times are measured with respect to reference frames, not with respect to people or other localized objects.
You'll have a much easier time if you don't conflate localized objects and reference frames. If someone tells you that an object A is "in" an inertial frame S, pretend they said that A is at rest wrt S, since that's what they meant.
If I understand your second question correctly, the times measured by A and B will be 1 and 2 seconds respectively. They are different measurements, referred to different coordinate systems, so there's no reason for them to be equal. The second postulate is about the speed of light wrt inertial frames, not the speed of light relative to other localized objects.
