Symmetric means that the source group and all the intermediate groups are the same, i.e. $\mathbb{G}_1 = \mathbb{G}_2 = ... = \mathbb{G}_{n-1}$ using the notation of the paper you linked, while asymmetric means that they can be different.
Check, for instance, remark 1 on the original paper about multilinear maps.
And about the vectors as indices, the aim is just to identify the levels more clearly. Since you have $2^n$ levels, you could use numbers from $0$ to $2^n-1$ to identify them, but then, increasing the levels with the multiplication would not have this nice interpretation of adding the entries of the vectors (or, equivalently, doing a union of the sets that represent the levels).
In several papers, for the symmetric scenario, we have a ring $R$ and an invertible element $z \in R$. Then, the map goes from $(z^{-1}R)^n$ to $z^{-n}R$. But, since the schemes are actually graded, you can apply the map partially and get elements in $z^{-k}R$ for $k < n$. Then, this $k$ is the level.
So, basically, we have elements of the form $r_i / z$ and you can multiply $n$ of them to get $(\prod_{i=1}^nr_i) / z^n \in z^{-n}R$.
In the asymmetric scenario, instead of a single $z$, we have $z_1$, .., $z_n$. Then, the map goes from $z_1^{-1}R \times ... \times z_n^{-1}R$ to $(z_1^{-1}\cdots z_n^{-1})R$.
So, you have elements $r_j / z_i$ and you can multiply them. But now, you can get any product of $z_i$'s in the denominator, so, it is easier to represent the levels by the exponents of each $z_i$, or, equivalently, with an $n$-dimensional integer vector.
For example, considering $n = 4$, an element $r / (z_1 z_3)$ is in level $(1, 0, 1, 0)$, and an element $r' / z_4$ is in level $(0, 0, 0, 1)$. Therefore, combining them (applying the map partially), gives us $r r' / (z_1 z_3 z_4)$ that is in level $(1, 0, 1, 1)$.
Thus, the vectors used to represent the levels tell us which groups were already used in evaluation of the map.
