Tag badges per user SEDE query
Implementation

The query is quite well written for what it does.  I just have a few minor nitpicks about its implementation.

The FROM, GROUP BY, and ORDER BY clauses are all subsidiary to the SELECT clause.  Therefore, I would indent them by one level.  (Indenting your INNER JOIN clauses so that they are part of the FROM clause was good.)

You're getting the disadvantages of the comma-first convention without reaping the benefits.  The advantage of putting commas first is that it makes it foolproof to add or remove columns in the SELECT: you should be able to delete an entire line of text without having to worry about fixing commas on neighbouring lines.  However, to take advantage of that property, you should put the first column on the same line as the SELECT keyword:

SELECT Users.Id as [User Link]
     , Users.DisplayName
     , COUNT (Badges.Id) AS Number_of_Tag_Badges
    FROM …


Remove the space before the parenthesis in the COUNT (…) call.

The joins would be better if the tables were listed linearly:

$$ \fbox{Users} \overset{\texttt{UserId}}{\longleftrightarrow}
   \fbox{Badges} \overset{\texttt{Name}}{\longleftrightarrow} \fbox{Tags}$$

Since the displayed results are user-centric, I would suggest listing the tables as Users, Badges, Tags rather than any other order.

Concept

It would be handy if you added a rank column.

The query doesn't distinguish between gold, silver, and bronze badges.  You should be able to report the breakdown.  The convention used by the International Olympic Committee is to sort by the gold medal count first, rather than by the total.  Perhaps you should do the same.

WITH EarnedTagBadges AS (
    SELECT Users.DisplayName
         , Badges.UserId
         , Badges.Name
         , RANK() OVER (PARTITION BY Users.Id, Badges.Name ORDER BY Date) AS Level
        FROM Users
            INNER JOIN Badges ON Users.Id = Badges.UserId
            INNER JOIN Tags ON Badges.Name = Tags.TagName
), UserTagBadges AS (
    SELECT UserId
         , DisplayName
         , COUNT(CASE WHEN Level = 3 THEN 1 END) AS Gold
         , COUNT(CASE WHEN Level = 2 THEN 1 END) AS Silver
         , COUNT(CASE WHEN Level = 1 THEN 1 END) AS Bronze
         , COUNT(Level) AS Total
        FROM EarnedTagBadges
    GROUP BY UserId, DisplayName
)
SELECT RANK() OVER (ORDER BY Gold DESC, Silver DESC, Bronze DESC) AS Rank
     , UserId AS [User Link]
     , DisplayName
     , Gold
     , Silver
     , Bronze
     , Total
    FROM UserTagBadges
    ORDER BY Gold DESC, Silver DESC, Bronze DESC;