It is possible to count the number of non-crossing diagrams including $k$ bridges by using Catalan numbers. For instance in Fig 1, I show the 5 possible diagrams obtained for $k = 3$ bridges, which equals $Cat(3)$.
In general $$ N_{n.c.} (k) = Cat(k) $$ with $N_{n.c.}(k)$ the number of non-crossing diagrams with $k$ bridges.
My question is: is there a way to count the number of diagrams with a fixed number $\alpha$ of crossings: $N(k, \alpha)$? For instance, for $k=2$, we have $$ N(2, 0) = N_{n.c.}(2) = 2\;, \qquad N(2,1) = 1 $$ which are represented in Fig2. Note that I exclude degenerate cases where more than two bridges cross at the same point. Also starting and ending points of different bridges cannot coincide.
The total contributions from diagrams with an arbitrary number of crossings $\alpha$ is also easy to compute: $$ N_{tot}(k) = \sum_{\alpha\geq 0} N(k, \alpha) = \frac{(2k)!}{2^k k!} $$
But what about the single coefficients $N(k, \alpha)$?
In 2002 I gave an expression in alt.math.recreational, translated to your $k$ and $\alpha$ as
$$N(k,\alpha)=\sum_{j} (-1)^j { (k-j)(k-j+1)/2-1-\alpha \choose k-1} \left({2k \choose j}-{2k \choose j-1}\right)$$
where $j$ is such that $0 \le j \le k$ and $(k-j)(k-j+1) \ge 2(k + \alpha)$
OEIS A067311 has more information, as this is equivalent to counting handshakes across a table and the number of crossings as illustrated in