You should divide 20Gbps by 9 bits (2275 MB/s) or even by 10 (2048 MB/s), not by 8, because in addition to useful payload, there are also packet headers and control packets being transferred. So it actually doesn't.
"Packet headers and control packets" are not the reason you should divide by 10. The reason is that Thunderbolt uses 8b-10b encoding so 1 byte of data is transferred as 10 physical bits. Therefore the maximum theoretical usable bandwidth is 20e9/10/1024/1024 = 1907 MiB/sec. Then on top of that you have to deduct the overhead from packet headers and control packets so the real-world usable bandwidth is even less than 1970 MiB/sec...
Doesn't TB2 have two interleaved streams, one carrying DP1.2 and the other carrying 20Gbps PCI-E? Even if the PCI-E stream has enough bandwidth for 5K in theory, the ecosystem isn't in place to use it as a display output. That's what the DP part is for.
For other people to play with this, just google "(5120 * 2880 * 3 * 60) bytes/s in Gib/s" to get the answer (just short of TB2s 20GiB/s assuming no overheads.
14745600pixels x 3 colours, 1 byte each = 44236800 bytes
44236800/1024/1024 = 42.19MB each frame
42.19MB per frame x 60 frames a second = 2531.25MB.
TB2 has a bandwidth of 20Gbps = 2560MB/s
So yeah, theoretically TB2 has enough bandwidth. But it's a very tight fit.