A path in the unit square that "doubles back" on itself in a nice way

This is just about the asymptotic behavior for $a\to+\infty$. I claim that the minimal length $\ell$ is about $\sqrt{2a}$ for large $a$.

The upper bound

Consider $2n$ horizontal lines splitting the square into strips of width $\frac 1{2n-1}$ Now travel along the odd-numbered lines (in the natural enumeration from the top) in the natural way: left to right on line 1, down to line 3, right to left on line 3, doun to line 5, left to right on line 5, and so on. Then return to line 2 in constant time and repeat the process going over even lines in the same fashion. Notice that every strip is bounded by one odd line and one even line and the time between going over those lines is about $n$ (with constant error). So, if $x$ is in the strip, we can drop off at the nearest point on the odd line and get picked up at the nearest point on the even line walking the total distance $\frac 1{2n-1}$. Thus if $\frac 1{2n-1}a<n-C$, we have the desired property, i.e., we can take $n\approx \sqrt{a/2}$ for large $a$ giving the length $\approx 2n\approx \sqrt{2a}$.

The lower bound

Assume we have an admissible curve $\gamma$ of length $\ell$. Parameterize it by the arc length $t\in[0,\ell]$. Suppose we have any (reasonably decent) function $f(t)$ with the property $f(t')+f(t'')\ge |t'-t''|/a$. Then every point $x$ must lie in the union of disks $D_t$ centered at $\gamma(t)$ of radius $f(t)$ (if we can reach $x$ by getting out at $t'$ and getting back on at $t''$, then, since we walked at most $|t''-t'|/a$, either $x$ is $f(t')$ close to $\gamma(t')$, or $x$ is $f(t'')$ close to $\gamma(t'')$). Thus the area of the union of those disks should be at least $1$. On the other hand, under some mild smallness and regularity assumptions on $f$, we can bound it by $2\int_0^\ell f(t)\,dt+o(1)$. Now just use $f(t)=a^{-1}|t-\frac\ell 2|$ to get the bound $\ell^2/2\ge a(1+o(1))$, so $\sqrt{2a}$ cannot be improved much for large $a$.

This is not an answer and adds little, but ... It maybe easier to consider a surrounding disk rather than a square. I like the OP's idea of a spiral. Concentric circles allow $a>1$ shortcuts:

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          ^{$2 \sin \theta / 2 < \theta$.}

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A tight spiral approximates concentric circles.