This is a gaps and islands problem, You can solve it using window function sum() to calculate a running total, this creates islands where rows with consecutive months have the same rn value.

The max() window function used to find the maximum amount within each island for each account.

with cte as (
  select *, sum(Amount) over (partition by Account order by MONTH(Period + ' 1 2014') ) as rn
  from TableA
)
select Account, Period, max(Amount) over (partition by Account, rn ) as Amount
from cte

Results :

Account Period    Amount
AC100   January   100
AC100   February  100
AC100   March     100
AC100   April     100
AC100   May       100
AC100   june      600
AC100   july      700
AC100   August    700
AC100   September 700
AC100   October   1000
AC100   November  1000
AC100   December  1200

Demo here

Using your example data:

create table #data
(
    Account varchar(5),
    [Period] varchar(9),
    amount int
)

insert into #data values
('AC100','January',100),
('AC100','February',0),
('AC100','March',0),
('AC100','April',0),
('AC100','May',0),
('AC100','June',600),
('AC100','July',700),
('AC100','August',0),
('AC100','September',0),
('AC100','October',1000),
('AC100','November',0),
('AC100','December',1200)

You can achieve this with an OUTER APPLY to find the last non-zero amount and return that instead when the amount value is zero:

select
    d.Account,
    d.[Period],
    case when d.amount = 0 then lastNum.lastAmount else d.amount end as amount
from #data d
outer apply
(
    select top 1 amount as lastAmount
    from #data l
    where month(l.[Period] + '1 1900') < month(d.[Period] + '1 1900')
    and l.amount <> 0
    order by month(l.[Period] + '1 1900') desc
) lastNum

Results: