Why does using double brackets "[[" to check if wildcard matched files exists fail, while single brackets "[" work?

The -e conditional operator tests whether a file with the given name exists. It doesn't do any pattern matching. [[ -e fileclass1*1.txt ]] tests whether there is a file called fileclass1*1.txt with a literal asterisk in the name, not whether there is at least one file matching the wildcard pattern fileclass1*1.txt.

With a single bracket [ ! -e $name ], since $name is unquoted, the value of name undergoes wildcard substitution and word splitting. If the value of name is Directory1/fileclass1*1.txt, the effect depends on how many files match the wildcard pattern.

• If the pattern doesn't match any file, it's left unexpanded. Then the -e operator sees the file name Directory1/fileclass1*1.txt, which doesn't exist, so [ ! -e $name ] is true.
• If the pattern matches a single file, it's replaced by that file's name. Then the -e operator sees that file name, which exists¹, so [ ! -e $name ] is false.
• If the pattern matches two or more files, it's replaced by the list of matching file names, which results in a syntax error in the conditional. In this case, [ ! -e $name ] displays an error message and returns a failure status.

The effect is different with [[ -e $name ]] because [[ … ]] is special syntax (whereas [ is an ordinary command whose arguments are expanded normally). No word splitting or wildcard expansion is done inside [[ … ]].

To solve your actual problem, see Test if there are files matching a pattern in order to execute a script

¹ _{Except in the unlikely case that due to a race condition, the file is removed between the time the shell lists the directory content to expand the wildcard pattern, and the time the -e operator checks for the file's existence.}