(Sorry for not answering this on the previous post, you asked this question before. By the way, you didn't say which paper you are reading...)
$Map(\mathbb R^{0|0},M)=M$. Trivial.
$Map(\mathbb R^{0|1},M)=\Pi T M$. ($TM$ is the tangent bundle of $M$, and $\Pi$ reverses grading of a vector bundle. So $\Pi T M$ denotes the total space of the degree-reversed tangent bundle of $M$. If $\dim M=d|\delta$, then $\dim \Pi T M=d+\delta|d+\delta$.)
Since
$Hom(S, Map(\mathbb R^{0|n},M))
= Hom(S\times \mathbb R^{0|n},M)$
$= Hom(S\times \mathbb R^{0|n-1}\times \mathbb R^{0|1},M)
= Hom(S\times \mathbb R^{0|n-1},\Pi T M)$,
we obtain inductively $Map(\mathbb R^{0|n},M)=(\Pi T)^n M$.
However, the interesting thing about $Map(\mathbb R^{0|n},M)$ is that it has an action by
$Diff(\mathbb R^{0|n})$, the supermanifold of invertible maps from $\mathbb R^{0|n}$ to itself.
I guess this is not so visible if you write $(\Pi T)^n M$, since this was obtained by destroying the symmetry in the odd coordinates.
For a description which keeps the symmetry:
$Map(\mathbb R^{0|2},M)$ is the pullback of $\Pi( T M\oplus T M) \to M$ along $TM\to M$,
which I learned from Dan Berwick Evans at Berkeley.
I would guess this is as explicit as it gets in general, and that probably more difficult pullbacks squares involving $\Pi T M$ exist for $Map(\mathbb R^{0|n},M)$ with bigger $n$.
One can find a description and discussion of $Map(\mathbb R^{0|n},M)$ in the paper with the nice title "Differential gorms, differential worms", Denis Kochan, Pavol Severa arXiv:math/0307303.
