There are no continuous bijections from $\mathbb{R}$ to $\mathbb{R}^2$, or to $[0,1]^2$.
Suppose $f$ is a continuous injection from $\mathbb{R}$ into $\mathbb{R}^2$. Then for each $n \in \mathbb{N}$, $f|_{[-n,n]}$ is a continuous injection from a compact space to a Hausdorff space, and hence a homeomorphism onto its image. Thus the image $f([-n,n])$ is nowhere dense: it's compact, hence closed, hence if it were somewhere dense it would contain a closed ball. But then there would be infinitely many points that could be deleted from it without disconnecting it, contradicting it being homeomorphic to $[-n,n]$.
Thus the image of $f$ is a countable union of nowhere-dense sets. So by the Baire category theorem it can't be all of $\mathbb{R}^2$, or indeed any set with nonempty interior.
The first problem is that "distance along the curve" is not meaningful for a space-filling curve -- the usual definition of distance for smooth curves (limit of approximating polygons) leads to the distance between two general points on the curve being infinite.
If you replace "distance along the curve" with "difference in parameters", then I can do it for $[0,\infty)\to\mathbb R^2$, but getting a doubly-infinite domain seems to be impossible.
Possible for $[0,\infty)\to \mathbb R^2$: Divide the plane into unit squares and number them in a spiral going out from the origin:
12 11 10 9 .
13 2 1 8 .
14 3 0 7 .
15 4 5 6 21
16 17 18 19 20
The classic Peano curve construction yields a space-filling curve $[0,1]\to[0,1]^2$. Put countably many of these together into $f:\mathbb [0,\infty)\to\mathbb R^2$, such that $f([n,n+1/2])$ fills the square numbered $n$ in the diagram, and $f([n+1/2,n+1])$ is just a simple connection from the endpoint of one Peano curve and the begining of the next.
This $f$ fills space but does not satisfy distance locality. But $g(t) = f(t^3)$ does. The cubic growth ensures that each winding in the spiral occupies a bounded amount of $t$, so the parameter distance between each square and any of its 8 neighbors is globally bounded.
Impossible for $\mathbb R\to \mathbb R^2$: Assume $f:\mathbb R\to\mathbb R^2$ is space-filling and satisfies your distance locality criterion. With appropriate scaling and translation we can assume $f(0)=(0,0)$ and $a=b=1$. We can then prove by induction on $n$ that $|f(t)|<n \Rightarrow |t|<n$ for all integers $n$, and therefore $|f(t)|>|t|-1$ in general. Then let $D=1+\max\limits_{|t|\le 1}|f(t)|$, and consider the two sets
$$ P = \{ f(t) \mid t > 1 \land |f(t)| > D \} $$
$$ M = \{ f(t) \mid t < -1 \land |f(t)| > D \} $$
$P$ and $M$ are both nonempty -- one contains $f(D+1)$, the other $f(-(D+1))$ -- and their union is the connected set $\{x\in \mathbb R^2\mid |x|\ge D\}$. Therefore their closures must intersect. But near a point where the closures meet, there must be $t$s that violate distance locality.
Best Answer
The images you provided don't lead to the construction of a space-filling curve. This is due to the fact that if you take $\{f_n\}$ to be the sequence of functions defining the iterations that you have provided, normally we'd call $f:= \lim_{n\rightarrow \infty} f_n$ our resulting space-filling curve. However, in this case this limit doesn't exist (so $f$ isn't well-defined).