MySQL 8.0

By using db<>fiddle, you agree to license everything you submit by Creative Commons CC0.
create table t as
    select 1 as id, 1 as userid, 0  as first,  5 as amount union all
    select 2 as id, 1 as userid, 10 as first, 10 as amount union all
    select 3 as id, 1 as userid, 10 as first,  5 as amount union all
    select 4 as id, 1 as userid, 15 as first, 10 as amount union all
    select 5 as id, 2 as userid, 0  as first, 10 as amount union all
    select 6 as id, 2 as userid, 0  as first,  5 as amount union all
    select 7 as id, 2 as userid, 10 as first,  5 as amount  
select *
from t


id
userid
first
amount


1
1
0
5

2
1
10
10

3
1
10
5

4
1
15
10

5
2
0
10

6
2
0
5

7
2
10
5


      select t.*
        ,first + amount as last -- last value in range
        ,( -- maximum end of all previous rows
           select max(first + amount) 
           from t as t2
           where t2.userid = t.userid
             and t2.first < t.first
         ) as prevlast
      from t


id
userid
first
amount
last
prevlast


1
1
0
5
5
null

2
1
10
10
20
5

3
1
10
5
15
5

4
1
15
10
25
20

5
2
0
10
10
null

6
2
0
5
5
null

7
2
10
5
15
10


   select userid
     ,prevlast
     ,min(first) as minfirst -- first of group
     ,max(last) as maxlast   -- last of group
      -- if there was a gap, calculate length of gap
     ,min(case when prevlast < first then prevlast - first else 0 end) as gaplen
   from
    (
      select t.*
        ,first + amount as last -- last value in range
        ,( -- maximum end of all previous rows
           select max(first + amount) 
           from t as t2
           where t2.userid = t.userid
             and t2.first < t.first
         ) as prevlast
      from t
    ) as dt
   group by userid, prevlast


userid
prevlast
minfirst
maxlast
gaplen


1
null
0
5
0

1
5
10
20
-5

1
20
15
25
0

2
null
0
10
0

2
10
10
15
0


select userid
  -- overall length minus gaps
  ,max(maxlast)-min(minfirst) + sum(gaplen) as total
from
 ( 
   select userid
     ,prevlast
     ,min(first) as minfirst -- first of group
     ,max(last) as maxlast   -- last of group
      -- if there was a gap, calculate length of gap
     ,min(case when prevlast < first then prevlast - first else 0 end) as gaplen
   from
    (
      select t.*
        ,first + amount as last -- last value in range
        ,( -- maximum end of all previous rows
           select max(first + amount) 
           from t as t2
           where t2.userid = t.userid
             and t2.first < t.first
         ) as prevlast
      from t    ) as dt
   group by userid, prevlast
 ) as dt
group by userid
order by userid


userid
total


1
20

2
15


select userid, min(first), max(first + amount) as last
from (select t.*,
             sum(case when prev_last >= first then 0 else 1 end) over
                 (partition by userid order by first) as grp
      from (select t.*,
                   max(first + amount) over (partition by userid order by first range between unbounded preceding and 1 preceding) as prev_last
            from t
           ) t
     ) t
group by userid, grp;


userid
min(first)
last


1
0
5

1
10
25

2
0
15


with islands as (
      select userid, min(first) as first, max(first + amount) as last
      from (select t.*,
                   sum(case when prev_last >= first then 0 else 1 end) over
                       (partition by userid order by first) as grp
            from (select t.*,
                         max(first + amount) over (partition by userid order by first range between unbounded preceding and 1 preceding) as prev_last
                  from t
                 ) t
           ) t
      group by userid, grp
     )
select userid, sum(last - first) as total
from islands
group by userid;


userid
total


1
20

2
15
id	userid	first	amount	last	prevlast
1	1	0	5	5	null
2	1	10	10	20	5
3	1	10	5	15	5
4	1	15	10	25	20
5	2	0	10	10	null
6	2	0	5	5	null
7	2	10	5	15	10