I thought it gave it a global symbol a local value?

It does, at run-time.

• Foo.pm gets compiled.
• Foo.pm's topmost level gets executed.
  • local $Foo::bar gets executed.
    • $Foo::bar current value (undef) is saved.
    • $Foo::bar is set to undef.
  • Anything else in Foo.pm's topmost level gets executed.
  • $Foo::bar reverts to undef. A file counts as block for scoping purpose, and we've just run to the end of the block in which local $Foo::bar was called.
• &Foo::init is called.
• &Foo::init sets $Foo::bar to 42. There's no local $Foo::bar in effect, so the value remains.

What is the purpose of the declaring $Foo::bar local?

None, unless it's used in Foo.pm's topmost level, or in a function (directly or indirectly) called from Foo.pm's topmost level.

If so, when init is called will it set the localized value like a closure or the global value?

It depends on whether someone in the call stack did local $Foo::bar. Unless &Foo::init is called from Foo.pm's topmost level (where there is such a local), probably not.

Further, what is the point of qualifying the identifier with the package name? and does that have any effect on the functionality?

Removing it will result in a compilation failure because use strict 'vars' is active. If you want to remove the need for the package quantifier, use our $bar; in scope at compile-time.

use strict;
use warnings;

package Foo;

our $bar;  # "our" is like "my", but it
           # creates a package variable
           # instead of a lexical variable.

sub init {
    $bar = 42;  # "our" removes the need to
                # specify the package name.
}

...

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