What exactly is the torus $(\mathbb{Z}/n\mathbb{Z})^d$?
I know that $\mathbb{Z}/n\mathbb{Z}$ is the set of equivalence classes: $\mathbb{Z}, \mathbb{Z}+1, \mathbb{Z}+2, \ldots, \mathbb{Z}+(n-1)$. I also know that a simple torus is $S^{1} X S^{1}$ that is the cartesian product of 2 circles equipped with the product topology and the donut that we see is just the projection of this in the 3d space.
I think that $(\mathbb{Z}/n\mathbb{Z})^d$ is just a set of vectors with $d$ coordinates with each coordinate ranging from 0 to $n-1$but these points should be connected using edges or something like that. I don't know how this torus looks geometrically or even what this torus is exactly. Could you please give me an example of the image when $d=3$ and for a higher value of $d$?
Let's start with $d = 1$ and with $[0,n] \subset \Bbb{R}^1$. We can make a $1$-torus (a circle) out of this by identifying the two ends (the end at $0$ and the end at $n$). This gives us $[0,n]$ joined to make $\Bbb{R}/n\Bbb{R}$. Now if, we highlight the points of $\Bbb{Z} \subset \Bbb{R}$ in the original copy of $\Bbb{R}$, the integer points of $[0,n]$ light up, the lights at $0$ and $n$ are identified, and we end up with $n$ lit points on $\Bbb{R}/n\Bbb{R}$. These lights mark a complete residue system of $\Bbb{Z}/n\Bbb{Z}$.
You should be able to see the topological $1$-torus when we glue the ends of the interval together and see the discrete $\Bbb{Z}/n\Bbb{Z}$ structure when we light up the points. Note that adding or subtracting residues modulo $n$ corresponds to rotation along the circle $\Bbb{R}/n\Bbb{R}$ and if we start with two elements of $\Bbb{Z}/n\Bbb{Z}$, their addition (or subtraction) produces a point of $\Bbb{Z}/n\Bbb{Z}$.
For $d = 2$, start with $[0,n] \times [0,n] \subset \Bbb{R}^2$. Now we have "four ends", the four edges of the square. We identify the left and right along $(0,y) \sim (n,y)$ for $y \in [0,n]$. Topologically, we have a closed cylinder (without the disk endcaps, but including the two boundary circles). Now we identify the top and bottom ends along $(x,0) \sim (x,n)$ for $x \in [0,n]$. (We should be careful about the four points identified at $(0,0)$ since everywhere else we always identify two points at a time, and this is a good exercise for the reader.) Equivalently, we identify the two boundary circles in the same way. Note that we could have been "sneaky" and introduced a twist, e.g. $\{(0,a) \sim (n,a+1 \pmod{n}) \mid a \in [0,n]\}$. This would still give $(\Bbb{R}/n\Bbb{R})^2$ topologically, but the $(\Bbb{Z}/n\Bbb{Z})^2$ living in it would not add and subtract by translation parallel to the axes of grids of points; there'd be a little twist: $(0,n-1)+(0,1) = (1,0)$. (This would behave a little (but only a little) like a two digit, base $n$, representation of an integer -- adding $(0,1)$ would correspond to adding $1$ and adding $(1,0)$ would correspond to adding $10_n \pmod{n^2}$. However, multiplication would seem to be a random mess, not modeling integer multiplication modulo anything.)
Note that this torus prefers in embed in $\Bbb{R}^4$ (One circle, an $S^1$ likes to live in $\Bbb{R}^2$ and so does the second one, so the $2$-torus, $S^1 \times S^1$, embeds without distortion in $\Bbb{R}^4$.), where all the circles can have the same diameter simultaneously, so the addition/subtraction along the rows and columns is rigid rotation along a cirle. The usual "doughnut" or "bagel" embedding in $\Bbb{R}^3$ distorts the circles -- the circles around and near the "hole" are smaller than the circles around and far from the "hole". This means in $\Bbb{R}^3$, addition looks like rotations on one axis (around the hole) and like "rolling the surface of the torus" for addition/subtraction towards/away from the hole. This distortion is just a projection artifact from crushing our nice, regular torus into $\Bbb{R}^3$. But notice that we have cyclicity along both axes and independent addition in each component, so we do have the expected $(\Bbb{Z}/n\Bbb{Z})^2$ structure.
For $d = 3$, start with the cube $[0,n]^3 \subset \Bbb{R}^3$. Identify (the parallel faces without twist, which will also identify the parallel edges and identify all the vertices) \begin{align*} (x,y,0) &\sim (x,y,n), \qquad x,y \in [0,n]^2 \\ (x,0,z) &\sim (x,n,z), \qquad x,z \in [0,n]^2 \\ (0,y,z) &\sim (n,y,z), \qquad y,z \in [0,n]^2 \end{align*} Then $(\Bbb{Z}/n\Bbb{Z})^3$ is the image of $\Bbb{Z}^3$ in the original cube after the identifications. If you lay this $3$-torus out flat in $\Bbb{R}^6$, you can see the addition in each component is rotation along a circle and that these three additions are independent. If you stand inside this flat version and look in any of the (locally) $S^1$-parallel directions, you see a $\Bbb{Z}^2$ and another one one unit behind it and another one unit behind it and after $n$ of these, you see your back, and $n$ layers further, you see your back again and so on out to infinity. If you look at other angles, you either (rational slopes) see yourself some distance away or you don't (irrational slope) (assuming you are a point and you can somehow see and recognize points -- if you have positive volume, then every ray ends on one of the triply periodic images of you).
So how do you think about $d > 3$. Well, in detail, that's hard. In general, it's a $d-$dimensional cube with its $d-1$-dimensional faces identified (without twist), which identifies parallel $d-2$-dimensional faces, which identifies parallel $d-3$-dimensional faces, which ..., which identifies all the vertices (to one point). Then keep only the integral lattice living in that cube. Component addition is rotation along that component's circle, so we get $d$-independent addition directions, one per circle.