When you use the -> operator the '$' sigil is not necessary!

It has nothing to do with the -> operator. Few if any operators requires its operands to have a $ since they work on expressions, not specifically scalars.

For example, here are 4 different operators and not a sigil in sight:

foo()->{x}
foo() + bar()
!foo()
print(foo())
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Perl automatically understands the type i.e whether array or hash.

Correct.
EXPR->[...] expects EXPR to return an array reference.
EXPR->{...} expects EXPR to return a hash reference.
EXPR->(...) expects EXPR to return a code reference.

And also the variable is defined in the current package. Is that correct?

No. Symbolic references can access variables in any package.

>perl -le"$x='Foo::var'; ${$x}='abc'; print $Foo::var;"
abc
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And it happens because strict refs is off.

strict shouldbe used. It's there to prevent specifically these kinds of odd and hard to debug problems.