Probably the easiest way to see this is to use the formula for the curvature,
$$ \kappa = \frac{y'' x' - x''y'}{(x'^2+y'^2)^{3/2}}, $$
avoiding arc length like the plague (arc length integrals are in general basically impossible to work with, so it's better to just not bother normalising. Arc length only shows up in its derivatives anyway, which is why we can get away with this). Next we find the equation of the normal, which will be
$$ y-y(t) = -\frac{x'}{y'}(x-x(t)), $$
and then we set $(x,y)$ so that we are a distance $1/\kappa$ from the curve, which is the radius of curvature.
The relevant derivatives are
$$ x=t \qquad y = t^2 \\
x' = 1 \qquad y' = 2t \\
x'' = 0 \qquad y'' = 2,
$$
so the curvature is
$$ \kappa(t) = \frac{2-0}{(1+4t^2)^{3/2}}, $$
and the normal is
$$ (y-t^2) = -\frac{1}{2t}(x-t) $$
Then we also have $(y-t^2)^2 + (x-t)^2 = \frac{(1+4t^2)^{3}}{4}, $
and solving these simultaneously gives
$$ x = -4t^3, \qquad y = \frac{1}{2}+3t^2, $$
when you take the correct sign of the differences.
Alternative: go back to Cauchy's definition: find the intersection of two normals, then let one approach the other. Since we have the general normal, we can find the intersection as when
$$ 2(y-t^2) = -\frac{1}{2t}(x-t) = 1-\frac{x}{t} \\
2(y-s^2) = 1-\frac{x}{s} $$
Hence
$$ x = -2st(s+t), \qquad y = \frac{1}{2} + t^2+ts+s^2, $$
and we set $s=t$ to find
$$ x=-4t^3, y=\frac{1}{2}+3t^2. $$
And can I just say, I thought it was wonderful how easily using the definition made calculating this: I must bear it in mind in future.
Edit: Better idea for the first way: clearly the normal vector to the curve at $t$ is
$$ n = \frac{1}{(x'^2+y'^2)^{1/2}} (-y',x'), $$
and then you just find $(x,y) + n/\kappa$, to abuse notation somewhat. This gives the general evolute coordinate for $x=x(t)$, $y=y(t)$ as
$$ \left( x - y' \frac{x'^2+y'^2}{y''x'-x''y'} , y + x' \frac{x'^2+y'^2}{y''x'-x''y'} \right) $$
