What differences would it make to live in $T^3$, in $S^2 \times S^1$, in $\mathbb{R}P^3$, or in $S^3$?

To collect all this discussion into an answer, let's begin by noting that most of your intuitive observations are regarding the fundamental group of the spaces you are looking at. The fundamental group of a space is defined by specifying a basepoint (e.g., your position) and then looking at all the paths in the space that start and end at your specified basepoint. If the space is path-connected then it doesn't matter which basepoint you pick, as the fundamental group will be the same no matter where you look. Formally, a path from a point $A$ to a point $B$ inside of a space $X$ is defined to be a continuous function $f: [0,1] \to X$ such that $f(0) = A$ and $f(1) = B$. I hope you can see fairly clearly how this nicely captures our intuitive notion of what a path should be in rigorous, formal language. The fundamental group consists of all the paths from your chosen basepoint to itself, aka loops, except that two loops are considered to be the same if one can be continuously deformed into the other, by a process called homotopy. On the fundamental group we define a composition of paths which is different from the usual composition of functions, and concocted so that the resulting path is defined on the same domain, namely the unit interval $[0,1]$ - given two paths $f$, $g$ we define their composition $f*g$ to be the path $f*g: [0,1] \to X$ defined by $$(f*g)(x) = \begin{cases} f(2x) & 0 \le x \le \frac12 \\ g(2x - 1) & \frac12 < x < 1 \end{cases}$$ This rigorously captures the idea of "first do one loop, then do the other".

To connect all this to your intuitive pictures, let's observe that the dog's leash represents the path the dog has taken. When your dog returns to you after having taken a trip round the torus, you notice that the leash gets "caught up around the space"; there's no way you can yank the leash back except to have your dog walk back the way he came from. If instead the dog keeps on walking forward, possibly in some other direction, what you get is that the leash is "doubly caught-up" around the space. This corresponds to the composition of the two loops - the first one going around one direction, the second one going around the other.

$\mathbb{RP}^3$ has a rather curious fundamental group. It consists of exactly two kinds of loops: first, the trivial kind, i.e. the kind in which you can yank the leash all the way back without the dog having to retrace its path, and second, the kind in which the dog has walked around the space exactly once. This is the only nontrivial kind of loop in this space. If you compose this loop with itself, you get back the trivial loop. That is, if your dog's leash gets caught around the space after your dog has made his travels, and he continues on forward to make one more trip round the space, then the leash will actually unravel as you yank it back to you.

Now, per SteveD's comment, $\mathbb{RP}^3$ is orientable, but if you want to gain intuition about non-orientable spaces (like for example $\mathbb{RP}^2 \times S^1$), and specifically about what might happen to your dog as he took a trip around an orientation reversing path, here's an illustration: Rememember your dog's lazy right eye? Of course you do. Well, now it's a lazy left eye. More generally, all features of the dog that initially appeared to you to be on his right side would now appear to be on his left, and vice versa. Of course, from the dog's perspective, it's the orientation of the rest of the universe that's changed. Weird, right? You may ask, "Why is the dog reversed about his left and right features? What about the bottom or top features? A very wise man once discussed this question in a much livelier and more illuminating manner than I ever could.

Welp, I hope all this discussion helps solidify your intuitive understanding of 3-dimensional spaces. Please let me know if that clears everything up for you.