Calculating free disk space
In basic it is CALL DSKF(0) for drive 1, but I guess that's not what you are asking for 
If drive name is valid, then the only way above simple calculation could produce an incorrect result, is if that BDOS call produces incorrect output itself.
What's your doubt here? Whether that BDOS call can be relied upon to work as specified? 
In basic it is CALL DSKF(0) for drive 1, but I guess that's not what you are asking for
No, PRINT dskf(x), where x is the drive number (1=A, 2=B etc) if x is 0, then it is the current drive.
You are right, a little mistake it is indeed PRINT instead of CALL
There are two files on a disk (no hidden files): one file of 332 bytes and one file of 2.000 bytes.
Why the free space on disk + file size1 + file size2 doesn't match the free clusters size? What am I missing?
On FAT-formatted disks, space is allocated in multiples of the cluster size. In your example:
A file with size 0 takes up a directory entry, but no disk space (0 allocated clusters).
A file with size 332 occupies a single cluster (=1024 bytes in this case).
A file with size 2000 occupies 2 clusters (=2048 bytes in this case).
So if you start with a freshly formatted disk (713K free) and write 2 files of 332 and 2000 bytes, then 1+2=3 of those 713 clusters will be occupied, and 710 should be left. And that "710" is the number dskf(x) and BDOS function should return.
If not, probably means you've got some more (hidden?) files on disk, or the disk is corrupted in one way or another.
As for file totals, MSX-DOS may report something like 2332 bytes in 2 files, 710K left. So total of file sizes = info from directory entries added, free space = derived from remaining clusters.
Perhaps a little confusing, but a non-issue imho as long as you understand what's being reported & how it's counted.
And the other way around: if you have 10 files of 50 bytes each on a disk (=10 clusters occupied), would you want MSX-DOS to report that as "10KB in 10 files" ? More confusing than adding the lengths from directory entries, I think.
