I'll give it a try (clearly, comments aren't enough, and the references I found are too much).

Given the permutation $\pi$ of $\{1,2,\dots,2k\}$ one can build an orientable surface with boundary as follows.

Take a topological disk, say the rectangle $[0,2k+1]\times[0,1]$ in the complex plane, and consider its quotient by the relation which identifies for each $i=1,\dots,k$ a small interval $[a-\delta,a+\delta]$ ($0<\delta<1/2$) around point $a=\pi(2i-1)$ with interval $[b-\delta,b+\delta]$ around $b=\pi(2i)$, *with orientation reversed*, that is $a+x\simeq b-x$.

[added: this is reversed from my comments under the question : here we glue things on the *bottom* edge of the rectangle. This has the purely cosmetic effect of numbering points in the usual --counterclockwise-- orientation of the boundary].

The result is a compact, connected, *orientable* surface with boundary, which you can picture in $3$-space by protruding "nearly horizontal" rectangles from the identified intervals, making them meet only in their tips and the prescribed pair (I'm really sorry for not being able to draw that here, and being probably very bad at describing mental pictures too). The orientation comes from projection to the horizontal plane.

The resulting surface $S_\pi$ depends only on the partition of $\{1,2,\dots,2k\}$ into the pairs $\{\pi(2i-1),\pi(2i)\}$, $i=1,\dots,k$, that is on the $\pi$-conjugate $\sigma$ of the involution $\sigma_0=(12)(34)\dots(2k-1)(2k)$. There are $k!2^k$ different $\pi$'s for one $\sigma$ (this amounts to what @Ycor said in a comment).

Let $b$ be the number of boundary components of $S_\pi$. A book-keeping exercise
tells you that $b$ is the number of cycles of the permutation $i\mapsto \sigma(i)+1$
(or its inverse $i\mapsto\sigma(i-1)$, or any of their conjugates, like $i\mapsto\sigma(i)-1$, that's the same number).

Now the all powerful surface classification theorem tells us that this (compact, connected, orientable) $S_\pi$ is (homeomorphic to) an (orientable) surface of some genus $g$ with the interiors of $b$ disjoint closed discs removed.

Why that ? Well, glue $b$ closed discs to the boundary components of $S_\pi$, and you have a (compact, connected, orientable) surface *without boundary*, connected sum of $g$ tori, by classification. Then undo.

Hence the amazing fact : there is only one model for each pair $(g,b)$ !
[this should be a footnote] That this is amazing may easily be overlooked nowadays, but can perhaps better be sensed by the fact that *mapping class groups* of surfaces, that is the discrete groups of components of their autoequivalences (homeo/diffeo doesn't matter here) aren't yet understood. For instance they share many properties with subgroups of linear groups, but they are not known to be linear aside from small cases. Yet they are the means to recognize a surface when we meet one.[end footnote]

Now in an oriented manifold $M^n$, one can count algebraically intersections of submanifolds or *cycles* of complementary dimensions, as Poincaré tolds us. Either triangulate and refine or use Thom's transversality, but in the end it's linear algebra, counting intersections by comparing the orientation of the direct sum of tangent spaces with the ambient one. This gives rise to the integer valued *intersection product* on cycles of complementary dimensions $p,q=n-p$, which depends only on their homology classes and is $(-1)^{pq}$-symmetric.

Crucially, it is *non-degenerate* when the manifold is *closed* (and oriented), meaning that $H_p(M,\mathbb{Z})$ mod torsion is isomorphic to $Hom(H_{n-p}(M,\mathbb{Z}),\mathbb{Z})$ via the intersection product. This is part of so-called *Poincaré duality*, expressed more neatly nowadays by adding *co*homology to the picture (but we don't need it).

In our context of $1$-cycles on oriented surfaces, intersection is skew-symmetric.

If $S_g$ is a closed (=without boundary), connected oriented surface of genus $g$, $H_1(S_g,\mathbb{Z})$ is a free abelian group with a non-degenerate skew-symmetric form, with standard "symplectic" basis $(a_1,b_1,\dots,a_g,b_g)$, meaning

$$a_i\cdot b_j=\delta_{ij},$$ $$a_i\cdot a_j=b_i\cdot b_j = 0.$$

But when there are $b\geq 1$ boundary components, $H_1(S_{g,b},\mathbb{Z})\simeq \mathbb{Z}^{2g+b-1}$, mapping to $H_1(S_g,\mathbb{Z})\simeq \mathbb{Z}^{2g}$ with kernel of rank $b-1$, preserving (clearly) the intersection form. [added comment] This could be proved using the homology exact sequence of the pair $(S_g \supset S_{g,b})$, but here it is simpler to use homotopy invariance of homology plus the fact that for $b>0$, $S_{g,b}$ is homotopy equivalent to the graph with one vertex and $2g+b-1$ loops.
This is also relevant below.

Hence the intersection form is non-degenerate on $H_1(S_{g,b},\mathbb{Z})$ iff $b\leq 1$.

Concerning the surface $S_\pi$, it should be clear that it is homotopy equivalent to a graph with one vertex and $k$ loops, and
that letting $h_i$ denote the class of a closed curve going *counterclockwise* [NOTE: this is reversed from my comments under the question] through the handle resulting from gluing the intervals $\pi(2i-1)\pm\delta$ to $\pi(2i)\mp\delta$ (in reverse) that

- the $h_i$, $i=1,\dots, k$ form a basis of $H_1(S_\pi,\mathbb{Z})\simeq\mathbb{Z}^k$
- the matrix of their intersection products is the same as your skew-symmetric form on $V_\pi$

(maybe up to sign, I switched to the upper half plane hoping to correct for this).

Hope this helps.