Let $\Lambda:= \{m +ni : m,n \in \mathbb{Z} \}$ be a complex lattice, so that our fundamental period parallelogram (fpp) is just the unit square. Define two complex numbers to be equivalent if $z_1 - z_2 \in \Lambda$.
If a point $z=x+yi \in \mathbb{C}$ has finite order $n$ in $\mathbb{C}/\Lambda$ (assuming $x \not=0$), then the line intersecting the origin (which is also the bottom left corner of the fpp) with slope $\frac{y}{x}$ goes from the bottom to the top (if $\frac{y}{x}>1$; there are other cases too) of the fpp $n$ times, before reaching the top-right corner exactly and repeating. This means that $\frac{y}{x}$ is rational. On the torus corresponding to the fpp this means that the line wraps around the torus $n$ times if the point $z$ has order $n$. (See picture below.)
Question: If $z \in \mathbb{C}$, not purely imaginary, has infinite order, does this happen if and only if $\frac{y}{x}$ is irrational? Does the resulting line/curve on the fpp self-intersect? Does it eventually fill up the entire fpp? Or does it leave most of it empty? What about its image on the torus? Is this a space-filling curve? Or am I just being sloppy with my intuition here?
Note: This is based on section 2.7, "Torus as $\mathbb{C}/\Lambda$" in Algebraic Geometry: A Problem Solving Approach by Garrity et al.
The points on your line with slope $k$ are parametrized as $P(t):=t+kti+\Lambda$, $t\in\Bbb{R}$. If we have $P(t_1)=P(t_2)$ for some $t_1\neq t_2$, then we get the pair of equations $$ t_2=t_1+m,\qquad kt_2=kt_1+n $$ for some integers $m,n$. Clearly $m\neq0$, so plugging the first equation into the second yields $km=n$ and therefore $k=n/m$ is rational.
But
We see that for the $x$-coordinates in the above pair of equations to match we need $t_2\in t_1+\Bbb{Z}$. For a fixed $t_1$ this happens only for countably infinitely many choices of $t_2$. In other words, even when $k$ is irrational the line passes through only countably infinitely many points with a fixed $x$-coordinate. Also, the line only covers a set of measure zero, because it is a countable union of finite length line segments.
What your visual intuition is telling you is that when $k=y/x$ is irrational, the line will be dense on the torus. This is because for any $t_1$, the points $t_1+i k(t_1+m)+\Lambda$, with $m$ an arbitrary integer, are on the line. As $k$ is irrational, the image of the subgroup $k\Bbb{Z}$ is dense on $\Bbb{R}/\Bbb{Z}$. Actually if we assume a bit more we can even say the following. If the triple $\{1,x,y\}$ is linearly independent over $\Bbb{Q}$, then the set of points $n(x+iy)+\Lambda$ gotten with an integer parameter $n$ is already a dense subset of $\Bbb{C}/\Lambda$. This is an instance of the generalized Kronecker's theorem. On the elliptic curve this means that integer multiples of such a point form a dense set.