Order DateTime, then group by Date (ignoring Time) + other field

As discussed this is gaps/islands problem which just requires isolating the additional distinct groups, using a running count and subtracting a count partitioned by each group of rows:

with grp as (
    select Convert(date, [date]) [Date], operation, [count], 
        Row_Number() over(order by [date]) 
        - Row_Number() over(partition by operation, Convert(date, [date]) order by date) gp
    from t
)
select date, operation, Sum([count]) [Count]
from grp
group by [date], operation, gp
order by [date] desc, gp desc

DB Fiddle

I'd probably wind up using a CTE (there's probably a way to do it similar to how you've already tried though). Something along the lines of:

WITH x AS(
    SELECT CAST([Date] AS DATE) [Date],
           Operation,
           SUM(t.[Count]) as [Count]
    FROM [MyTable]
    GROUP BY CAST([Date] AS DATE),
             Operation )
SELECT x.[Date], 
       x.Operation,
       x.[Count]
FROM x AS x
ORDER BY x.[Date] desc,
         x.Operation;

Although; if you're going down the route of wanting to show something like:

X calls came before Y visits, then there were another z calls before the end of the day.

Then you'll need something more custom, like the solutions that were linked in in the comments.