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docs: style: the template-value grammar, from the reference

The table here was wrong and I published it from probes instead of reading
putout-script.md and @putout/compare, which document every value. Both docs
now link upstream rather than asserting from memory.

What was wrong: `__` was listed as matching nothing, and it is the value that
means "any node". `__a` is "any node, and linked" - its value carries into the
replacement - and the two are not interchangeable. Measured:

  'f(__a)' -> 'g(__a)'   gives g(1), the value carries
  'f(__)'  -> 'g(__)'    gives g(1), unlinked still fills in
  'f(__)'  -> 'g(__a)'   throws "Looks like template values not linked"
  'f(__a)' -> 'g(__b)'   throws "Looks like template values not linked"
  'f(__a__)' -> 'g(__a__)'  leaves f(1) unchanged, exit 0

That last is the trap: two underscores bind nothing, so the text is emitted as
written and the rule reports success having changed nothing. The rule is that a
replacement may reuse a name the key declared, and only those.

Also adds the values my probes never reached: __args, __object, __array,
__imports, __exports, __args__a for linked args, "__a" for any string literal
and /__a/ for any regexp.

My probes read the wrong node: a match visitor is handed the outer node, so
every one reported Identifier, and I generalised from that. The correction in
memory/putout-rules.md follows - include takes type names, where `__` is a type
like any other and not a catch-all - with the asymmetry spelled out.

Lint clean, 151 mcp tests pass.

1081 of 1081 branches covered (100.0%)

Branch coverage included in aggregate %.

6670 of 6670 relevant lines covered (100.0%)

61.48 hits per line