Let me suggest you a game to make it clear: Create the inverse Fourier that means construct a function $Y(t)$ that describes your sequence.
Point all your data versus time. Then start using the following function (in Mathematica or any other math program) (instead of complex expression this is easier to understand the connect to roots of unity):
$$y_i(t,k)= \sum_{n=1}^k \frac{1}{k} \cos\left(\frac{2 \pi (n-1) t}{k}\right)$$
(Discrete)Plot it and test! You may need to add a couple of different such functions $y_i(t,k)$ together (superposition) while every function with different $k$ and also multiplied perhaps with different amplitudes $a_i$ to get your function $Y(t)$ (like playing LEGO):
$$Y(t)=\sum_i a_i\; y_i(t,k)$$
This equation would as good as possible mimic (inverse Fourier) your sequence. If you do so you will understand that the transform is exactly the opposit way, and what its details value for you.
I did this quite often and works great to have sequences comprehensive.
Your first example and a period of $k=2$ for instance trivially:
$$Y(t)=y_1(t,2)= 9.81 \sum_{n=1}^2 \frac{1}{2} \cos\left(\frac{2 \pi (n-1) t}{2}\right)$$
PS: By the way $y_i(t,k)$ converges.
Additional explanation:
I hope I got the set up of your question correctly: You have discrete sequences of data like $\{9.81,...\}$ or $\{9.81,...,1.00,9.81...\}$ on the time line and transform these via Fourier to frequency domain.
Perhaps I understand what is your issue, and this is probably deeper going (frequency/amplitude uncertainty principle). I brought above an exercise (a game that should help) but lets try to answer your questions more directly regarding the values.
Can anybody tell me what result of discrete fourier transform means?
The result of the Fourier Transform as you will exercise from my above description will bring you only knowledge about the frequency composition of your data sequences. That means for example 1 the zero 0 of the Fourier transform tells you trivially that there is no superposition of any fundamental (eigenmode) periodic sequences with corresponding frequencies. This is why I wrote you $Y=y_1$. Your sequence is a periodic eigenmode of itself. Trivially your numbers of value 9.81, how many they might be (256), do not provide any additional information than they have in the sequence itself, because they are just the amplitude of your preiodic sequence which is its own eigenmode. So the Fourier transform tells you: there is a trivial case of frequency of eigenmode sequence and your numbers tell you there is trivial case of amplitudes - i.o.w NOTHING CHANGES in your sequence the behaviour is of no change. In other words you will learn here that the spectrum contains only one eigenmode and that is the sequence itself with a ground amplitude 9.81. Nothing more than this!
The second example is drammatic different. The amplitude $(1.00-9.81)$ (actually the difference) is carrying great information (as for example the location of a particle in physics, certain behaivourial change in the sequence/ accelleration...). However in order to precisely point this out you need a Fourier that contains a spectrum of frequencies. So remind the case of wave and particle in physics, when you want to locate the particle at a certain time/location (analogue to your $(1.00-9.81)$ signal at a certain time) that means precise amplitude, you would need all fundamental frquencies and waves/modes of the spectrum to sum them up and obtain the $(1.00-9.81)$ at a percise certain time point (this is why in my previous message I wrote as good as possible). In an abstract interpretation this superposition and sum up is implicitly done by your measurement instrument, and it points precisely to $(1.00-9.81)$ at a percise certain time point i a sequence. The problem however is that when you have obtained this way a precise $(1.00-9.81)$ at a certain time then you (your device) can not observe the frequencies, because you summed them up (implicitly) in the waves; in reverse if you perform a Fourier transform on the data sequence recorded, as you did, you obtain all the frequencies of the fundamental waves/modes composing your signal $(1.00-9.81)$ but at the cost that you can not more precisely fix the amplitude $(1.00-9.81)$ (of course in process of your decomposition the result is decomposed and scattered).
This is the uncertainty principle of any spectrum/Fourier. Unfortunately what happens in your thought scenario is that you want to perform a Fourier transform but ask for the meaning of the amplitudes ($(1.00-9.81)$). The answer is clear: The Fourier transform / spectrum of frequencies does not give you any information about the amplitude of the superposition.
Side note: At the same time keep into mind that the first example is the buttom line or callibration of the second example.
Annecdotic: The first example is an orchestra of one drum and itself repeating at a certain frquency and 9.81 pitch. The second could be an orchestra of the same drum playing at the same frequency and 9.81 pitch but in addition many instrumental players who are scilent all time except of precisely one instance of time, when playing one singular shot of 1.00 pitch.
For example 3 we learned already that the Fourier transform and the spectrum of frequencies do not give you any information about your absolute values in time domain (pitch)! But it gives you information on how often in a sequence you will change to sit position or vice versa. So if the spectrum shows more high frequencies bandwidth, that means you change often your position and if low frquencies bandwidth it shows you less change position. The Fourier transform will tell you how you behave, whether you accellerate more often up and down or less. It does not give you any information about the amplitudes (uncertainty paradox). Trivially the information about the amplitudes is already what you see in the time domain.
So before interpreting your data decide whether you observe in time domain (amplitudes and absolute value of your data) or frquency domain (frequency, behaviour, change).
The resultant vector of 256 values represents decomposition of acceleration values into what?
Into a spectrum of frequencies that is a characteristic fingerprint of your accellaration behaviour brought in that sequence of data record.
Hope this helps you to better understanding and is Answer to your Question. I still suggest you to do the exercise I proposed. Nothing is better than own experience.
Best Answer
Here are short observations for your question. I think I'll add detailed explanation later.
Why frequency graphs are square?
This comes from nature of multidimensional DFT. One-dimensional DFT is just an expansion of vector in basis of complex harmonics. Similar construction can be applied for 2D images, but here we should take something like a tensor product of basis 1D signals along "$x$" and "$y$" axis of matrices. That'll be the basis for the space of 2D images. If we have a vector of length $N$, then we have $N$ vectors in Fourier basis, each of them stands for different frequency. When we have a $N \times N$ image and we take a tensor product of basic signals, we'll have $N^2$ combinations of frequencies along "$x$" and "$y$". Each vector from the Fourier basis for an image can be determined by its frequencies, which means form the tuple $(j, k)$. Frequency image shows the absolute value of expansion coefficient before the vector with frequencies $(j, k)$. So, that's why it's a square; generally it should be rectangular.
Why are the graphs point symmetrical around the center?
It's the consequence of formulas that used for Fourier expansion. For 1D DFT, coefficients before $f_j$ and $f_{N-j}$ are such that they have the same absolute value (this is true for dot products with real vector, as shown further). This property is inherited by 2D DFT and you see that $(j, k)$, $(N-j, k)$, $(N-j, N-k)$ and $(j, N-k)$ have the same absolute value.
In image 3, why is it a dense grid with 2 bright lines?
This image has structure that is much closer to a tensor product of 1D signals than the previous. And that 1D signal can be represented with very few harmonics. It would be interesting to see, how size of square affects the number of harmonics with big coefficients.
ADDITION
Here I'll try to explain rigorously some things that vere mentioned earlier. I don't know how much you know about DFT so I'll try to cover as much as possible. So, if we have $N$-dimensional comlex vector space, the Fourier basis can be given by following formula: $$ F_m (n) = \frac{1}{\sqrt{N}} \exp{\frac{2\pi i \cdot m \cdot n}{N}},$$ where $m$ stands for the $m$-th basis vector and $n$ is a component index (both are integer and $0 \leqslant m,\; n \leqslant N-1$).
It is well known that Fourier basis is also an orthonormal basis with respect to standard dot product in $\mathbb{C}^n$, i.e. $\langle u, v \rangle = \sum \limits_{i} u_i \overline{v_i}$. So, then we are looking for expansion coefficients we can use dot product to find them. Here comes detailed explanation for second question. If we took a real vector $v$ and compare it's scalar product with $F_j$ and $F_{N-j}$:
$$ \langle v, f_j \rangle = \sum\limits_{p=0}^{N-1} v_p \overline{f_j(p)} = \sum\limits_{p=0}^{N-1} v_p \overline{\frac{1}{\sqrt{N}} \exp{\frac{2\pi i \cdot j \cdot p}{N}}} = \frac{1}{\sqrt{N}}\sum\limits_{p=0}^{N-1} v_p \exp{\left ( -\frac{2\pi i \cdot j \cdot p}{N} \right ) }$$
$$ \langle v, f_{N-j} \rangle = \sum\limits_{p=0}^{N-1} v_p \overline{f_{N-j}(p)} = \sum\limits_{p=0}^{N-1} v_p \overline{\frac{1}{\sqrt{N}} \exp{\frac{2\pi i \cdot (N-j) \cdot p}{N}}} = \frac{1}{\sqrt{N}}\sum\limits_{p=0}^{N-1} v_p \exp{\left ( \frac{2\pi i \cdot j \cdot p}{N} \right ) } $$
Since all coefficients of $v$ are real numbers we see that $\langle v, f_{N-j} \rangle = \overline{\langle v, f_{j} \rangle}$ and these dot products have the same absolute value.
Now I'll explain what I meant by tensor product of basis vectors in case of 2D images. By tensor product $u \otimes v$ I meant matrice $A$ which coefficient $A_{i, j}$ is equal to $u_i v_j$. Now we have two indices to locate the element of image and two indices for the frequencies. The formula for $(m, n)$ element of $(j, k)$ Fourier basis vector is following: $$ F_{j,k} (m,n) = \frac{1}{\sqrt{N_1}} \exp{\frac{2\pi i \cdot j \cdot m}{N_1}} \cdot \frac{1}{\sqrt{N_2}} \exp{\frac{2\pi i \cdot k \cdot n}{N_2}},$$ if we deal with $N_1 \times N_2$ matrices, or $$ F_{j,k} (m,n) = \frac{1}{N} \exp{\frac{2\pi i \cdot j \cdot m}{N}} \cdot \exp{\frac{2\pi i \cdot k \cdot n}{N}} = F_j(m) \cdot F_k(n)$$ in case of $N \times N$ matrice.
We also can apply a standard dot product to a space of 2D images if we put $$ \langle u, v \rangle = \sum\limits_{m=0}^{N-1} \sum\limits_{n=0}^{N-1} u(m, n) \overline{v(m,n)}.$$ This nicely agrees with embedding $\mathbb{C}^{N\times N}$ in $\mathbb{C}^{N^2}$. It also can be shown that $F_{k, j}$ also forms an orthonormal basis. Coefficients for expansion can be found with analogous formulas via dot product. As I said earlier the mentioned property for real images is inherited by 2D DFT.