You are right in that gravity did not change during data collection. You are a victim of uncertainty, which is a very important part of experimental physics. I'm sorry in advance for the "wall of text", and I hope that this clears up some confusion.
The problem is that $1.50$ may not be exactly $1.500000000...$. Because the numbers are provided rounded, they are not exact and you have lost information. Imagine that my watch can only report time to the nearest second and my measuring tape only reports to the nearest meter. If I measure a car's movement and it moves 2.3 meters in 0.8 seconds (true measurements, 2.875 m/s) I am required to round all my data to the nearest round number (2 meters in 1 second, 2.0 m/s). So, if I calculate a number based on my rounded data, it won't perfectly reflect reality because my data do not perfectly reflect reality. Even though your numbers are more precise than my example (accurate down to hundredths of a second and centimeters), there is still some amount of uncertainty in the numbers.
Feel free to skip to the "What it all means" section. The stuff after this is pretty dry and obtuse.
Quantitative Explanation
Note that I'm going to call displacement $x$ and time $t$, just for convenience.
Let's have a look at a subset of the data as an example:
Time (s) | Displacement (m)
1.50 | 3.09
2.00 | 6.60
You know the displacement to two decimal places-- that means that, for $t = 1.50$, $x$ could be anywhere from $3.085$ to $3.094999... \approx 3.095$, and it would still be okay to call it $3.09$, as long as you're rounding to that number of significant digits. Similarly, the time might not be exactly $1.5$! So, if you calculate anything based on those numbers, there's a certain amount of uncertainty in the result. Since there are two variables ($t$ and $x$), they can both vary at once. With the equation $v = (x_f - x_i)/ \Delta t = $, $v$ is biggest when $\Delta x$ is maximized and $\Delta t$ is minimized (dividing by a smaller number yields a larger number), and $v$ is smallest when $\Delta x$ is smallest and $\Delta t$ is biggest. We don't know the true, unrounded values of the two variables, so any possible combination of them is just as good as any other. Here's the range of possible velocities based on the above displacements:
$t = 1.5 \rightarrow 2.0$:
$$
v_{max} = \frac{6.605 - 3.085}{1.995 - 1.505} \approx 7.18 \, \mathrm{m/s}
$$
$$
v_{mid} = \frac{6.600 - 3.090}{2.000 - 1.500} = 7.02 \, \mathrm{m/s}
$$
$$
v_{min} = \frac{6.595 - 3.095}{2.005 - 1.495} \approx 6.86 \, \mathrm{m/s}
$$
Quite a range! Let's do it again for the next time so that we can get a range of accelerations:
$t = 2.0 \rightarrow 2.5$:
$$
v_{max} = \frac{9.695 - 6.595}{2.495 - 2.005} \approx 6.36\, \mathrm{m/s}
$$
$$
v_{mid} = \frac{9.690 - 6.600}{2.500 - 2.000} = 6.18 \, \mathrm{m/s}
$$
$$
v_{min} = \frac{9.685 - 6.605}{2.505 - 1.995} \approx 6.04 \, \mathrm{m/s}
$$
Again, quite a range. Now, to find the max,min accelerations possible for your data, you take the same approach. For $a_{max}$, divide the biggest possible $\Delta v$ by the smallest possible $\Delta t$, and the reverse for the min. For the above numbers, you get accelerations like so:
$$
a_{min} = \frac{6.36 - 6.86}{2.505 - 1.995} \approx -0.98 \, \mathrm{m/s}^2
$$
$$
a_{mid} = \frac{6.18 - 7.02}{2.5 - 2} = -1.68 \, \mathrm{m/s}^2
$$
$$
a_{max} = \frac{6.04 - 7.18}{2.495 - 2.005} \approx -2.33 \, \mathrm{m/s}^2
$$
where "min" and "max" are used (somewhat sloppily) to mean "greatest in magnitude", not the true meanings of "minimum" and "maximum". If I haven't messed up the numbers, the intermediate accelerations could be anywhere in the above range!
What it all means: Now you should be able to recognize that, because your accelerations are derived from numbers of limited certainty and they fall within the range given above, they cannot really be said to be different-- they are said to be "within uncertainty" of each other. It would be appropriate to take the average of all your values and call that the best guess you have of the acceleration. If you wanted to be a true scientist about it, you might have a go at calculating the uncertainty in acceleration, which is nontrivial, or you could go the lazy route (like a lot of scientists :)) and use something like the standard error of all your acceleration values. Either way, the end-all is that the acceleration didn't change during the data collection, it only looks like it did because you didn't measure position and time precisely enough :)
If you're interested in reducing the uncertainty, read on!
These numbers are not acceptable, the range is too big. The problem is that our uncertainties ($0.005 \mathrm{m}$ and $0.005 \mathrm{s}$) are very large compared to our numbers: we're using time steps of half a second, but we have an uncertainty of $2 \cdot 0.005 = 0.01$ in each variable (twice the uncertainty because you're taking the difference of two values). You can reduce the effect by evaluating the difference across more than one time step (ie find the acceleration from $t=0 \rightarrow 1$ instead of $0 \rightarrow 0.5$. This way, the change in time (and displacement) is bigger, but the uncertainty remains the same, so it impacts the result less!
Calibrate and validate. These are two words you need whenever you are doing calculations of physical parameters from measurement. You need to know how the device responds to known accelerations in order to determine whether the numbers you are getting are meaningful at the level you want to use them. The easiest would be if the manufacturer provided you with such details. Sadly, these days, devices of this type often come with no documentation whatever.
The fact you are getting 9.3 to 9.5 for gravity is suggestive. It suggests you have either an offset or a scale issue. One way to check this would be to read the values with the device sitting on a table, then turn it upside down and read again. An offset will show as a difference up-to-down. A scale issue will show low values both ways.
If you have a way to put the device in a situation with a known acceleration that would help also. Maybe a turn-table or some such? Though some such devices might be fooled by spinning in too tight a circle.
The steady drift suggests you could have a response issue. It may not be linear with acceleration. A narrow-but-high spike may be clipped, for example. Testing for such gets pretty complicated. You would need known acceleration profiles and then see how the device responds. Correcting for it adds more complication still.
Best Answer
That the numerical value of the resulting acceleration is greater upwards than downwards is likely due to friction. Uphill friction and gravity pull in the same direction, while downhill friction and gravity pull in opposite directions.
You can use the difference in acceleration to estimate the friction. The average is the (projected) gravitational acceleration, from which you can calculate the inclination angle (assuming a constant slope).