How to reconcile time dilation in relativity with what you actually see? 
If a traveler is moving at $0.5c$ towards a clock which is located one light year away, his relativistic time dilation is $1.15$. But in the time he takes to arrive at the clock, he must catch up the 1-year lag that he saw when the clock was 1 light year away, in 2 years travel time. This means he must see the clock moving at twice its normal speed, not $1.15$.

What am I doing wrong?
Why was this question marked as "unclear"? I don't know how else to ask it.
If this question is so unclear, then how is it that I got three perfectly clear answers?
Anyway, thanks to those who took the time to answer.
Apparently, what I was doing wrong was to ask what I was doing wrong? Go figure.
No, this isn't my homework, as I have not attended any university since 1999.
 A: I suspect that you're trying to calculate this by multiplying time intervals by the Lorentz factor, $\gamma$. And you're ignoring length contraction. To do this properly, you need to use the full Lorentz transformations.
Don't mix up what each observer measures & calculates with what they observe; observations include the time delay due to the finite speed of light. Determine what are the important events in your scenario, and calculate their time & space coordinates in each frame.
BTW, if two frames $S_1$ and $S_2$ have a relative speed of $v$, so
$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
then each frame measures the other's clock to be ticking slower by the $\gamma$ factor.
A: There are three separate physical effects that relate directly to the passage or perception of time in SR, and almost every misconception or so-called paradox of SR arises where people have either confused the three effects or overlooked one or more of them. The three effects are as follows:
Time dilation. This arises where one clock moves between two other synchronised clocks that are stationary relative to each other. The passage of time on the moving clock is always less than the difference in readings between the first clock it has passed and the second.
The relativity of simultaneity. This is time dilation viewed from a different perspective. As a moving clock passes a series of stationary clocks which are synchronised in their own frame, the clocks will appear to be increasingly out of synch to the clock that is passing them.
The Doppler effect This arises where a clock is being viewed (or sending signals to) a distant observer who is moving relative to it. If the observer moves towards the clock, it will be 'blue-shifted' which means that it will appear to tick more frequently- if the observer is moving away, the clock will appear to be ticking more slowly.
These three effects all need to be taken into account, and distinguished so that the impact of one is not confused for the impact of another.
In the example you give, all three effects are at play. Suppose you had a friend, moving at the same speed as you and with a synchronised clock, who was present at the distant clock at the time you started your countdown. Likewise, suppose the owner of the distance clock had a friend who was stationary relative to it and standing next to you when you began your journey. Then the effects will work as follows:
The distant clock will seem to you, as you approach it, to be ticking more quickly than once per second, owing to the Doppler effect.
When you arrive at the distant clock, the time that has passed on your own clock will seem less than the time recorded for your journey by the owner of the distant clock and his friend who was present at your departure.
Your friend, who was at the distant clock when you set out, will tell you that the distant clock was showing the wrong time at the outset.
In the example you cite, you must use the relativity of simultaneity to work out what the time is in your frame at the distant clock when you start your journey, and you must consider Doppler effects to determine how the frequency of the distant clock will appear to be running fast as you approach it.
A: You are being downvoted unfairly.  Your answer is logically OK (in the sense that there is more to it than just the Lorentz Transform), but the calculation is wrong.  Use the doppler effect, and you will see the clock appearing to move $\sqrt {1.5 / 0.5}$ times (1.73) normal speed as you approach (it will also appear to run slower - by a factor of $\sqrt {0.5 / 1.5}$ - if you recede at 0.5$c$).
