It runs because by default executable file is assumed to be /bin/sh script. I.e. if you didn't specify any particular shell - it is #!/bin/sh.
The // is just ignored in paths - you can consider is at as single '/'.
So you can consider that you have shell script with first line:
/usr/bin/env go run $0 $@ ; exit
What does this line do? It runs 'env' with paramenters 'go run $0 $@'. there 'go' is command and 'run $0 $@' are args and exits script afterwards. $0 is this script name. $@ are original script arguments. So this line runs go which runs this script with it's arguments
There are quite interesting details, as pointed in comments, that two slashes are implementation-defined, and this script would become POSIX-correct if it specify three or more slashes. Refer to http://pubs.opengroup.org/onlinepubs/9699919799/basedefs/V1_chap04.html for details on how slashes should be handled in paths.
Note also that there is another mistake in script the $@ it's correct to use "$@" instead, because otherwise if any parameter contains spaces it will be split to many parameters. For example you can't pass file name with spaces if you not using the "$@"
This particular script obviously rely on the idea that '//' is equal to '/'