Running total for grouped data

you can just use a simple group by and computed order by for month:

 select Month, Region, Store, sum(Sales) as Sales
 from Sales
 group by Month, Region, Store
 order by charIndex (Month, 'Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec'), Region, Store

As Viorel pointed out, it gets less messy with numeric values for months. And presumably you need the year somewhere as well. Furthermore, I assume that you have tables that define the Regions and Stores. Those tables should replace the CTE AllStores.

An issue with Viorel's query is that performance will be proportional to the square of number of months. This solution avoids this issue:

DROP TABLE IF EXISTS #Peter
CREATE TABLE #Peter (Month char(3) NOT NULL,
 Region char(2) NOT NULL,
 Store char(2) NOT NULL,
 Sales decimal(10,2) NOT NULL
)
INSERT #Peter(Month, Region, Store, Sales)
VALUES
('Jan', 'R1',	'S1', 10.00),
('Jan', 'R1',	'S1', 15.00),
('Jan', 'R1',	'S1', 30.00),
('Jan', 'R1',	'S2', 5.00),
('Jan', 'R1',	'S2', 1.00),
('Jan', 'R2',	'S3', 17.00),
('Jan', 'R2',	'S3', 5.00),
('Jan', 'R2',	'S4', 60.00),
('Jan', 'R2',	'S4', 45.00),
('Jan', 'R2',	'S4', 11.00),
('Feb', 'R1',	'S1', 20.00),
('Feb', 'R1',	'S1', 40.00),
('Feb', 'R1',	'S1', 60.00),
('Feb', 'R2',	'S3', 10.00),
('Feb', 'R2',	'S3', 10.00),
('Feb', 'R2',	'S3', 10.00),
('Feb', 'R2',	'S3', 10.00)
go
; WITH AllStores AS (
	SELECT DISTINCT Region, Store FROM #Peter
), Months AS (
 SELECT value AS monthno, 
	 choose(value, 'Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 
	 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec') AS Month
 FROM generate_series(1, 23)
)
SELECT m.Month, a.Region, a.Store, 
 SUM(SUM(isnull(p.Sales, 0))) OVER(PARTITION BY a.Region, a.Store 
		 ORDER BY m.monthno
									 ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)
FROM Months m
CROSS JOIN AllStores a
LEFT JOIN #Peter p ON p.Month = m.Month
 AND p.Region = a.Region
						 AND p.Store = a.Store
GROUP BY m.Month, m.monthno, a.Region, a.Store
ORDER BY m.monthno, a.Region, a.Store
go
;WITH MonthlyStoreSales AS
(
 SELECT
 [Month],
 [Region],
 [Store],
 SUM([Sales]) AS MonthlySales
 FROM #Peter
 GROUP BY [Month], [Region], [Store]
)
SELECT
 [Month],
 [Region],
 [Store],
 SUM(MonthlySales) OVER
 (
 PARTITION BY [Region], [Store]
 ORDER BY [Month]
 ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
 ) AS Sales
FROM MonthlyStoreSales
ORDER BY [Month], [Region], [Store];

I deleted the AI answer, because it was not producing the desired result.

You could also try a longer query:

;
with Q0 as
(
	select *, charindex([Month], 'JanFebMarAprMayJunJulAugSepOctNovDec') as MonthIndex
	from MyTable
),
Q1 as
(
	select distinct Region, Store
	from Q0
),
Q2 as
(
	select distinct [Month], MonthIndex, Q1.Region, Q1.Store
	from Q0, Q1
)
select [Month], Region, Store, (select sum(Sales) from Q0 where Region=Q2.Region and Store=Q2.Store and MonthIndex<=Q2.MonthIndex) as Sales
from Q2
order by MonthIndex

It would be simpler if the month were numeric.