Most pythonic solution I could come up with :)

marksheet = []
for _ in range(0,int(input())):
    marksheet.append([input(), float(input())])

second_highest = sorted(list(set([marks for name, marks in marksheet])))[1]
print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest]))

The simplest and accurate solution:

Whith explaination of each line of code:

@python 2.7

d={} #1
for _ in range(int(raw_input())): #2
    Name=raw_input() #3
    Grade=float(raw_input()) #4
    d[Name]=Grade #5
v=d.values()#6
second=sorted(list(set(v)))[1] #7
second_lowest=[] #8
for key,value in d.items():  #9
    if value==second: #10
        second_lowest.append(key) #11
second_lowest.sort() #12
for name in second_lowest: #13
    print name #14

@1) Empty dictionary.

@2) Range for number of students.

@3) Accepting name of the student.

@4) Accepting the grade of the student.

@5) Assigning name as key and grade as value for the dictionary.

@6) Obtaining the values of dictionary(all grades of students.

@7) Remoing duplicte grades by using set data type , changing it to list, sorting in ascending order and taking the second lowest grade.

@8) Declaring an empty list for storing the name of the students who got the second lowest grade.

@9) Going through the name and grade of each students by using items() method of dictionary.

@10) Checking whether the grade is equal to the second lowest grade.

@11) If yes , append it to the second_lowest list.

@12) bSorting the name of students in asceding order

@13) Going through the name of each students who got the second lowes grade.

@14) Printing each name of students in seperate line.

List comprehensions are so great!

if __name__ == '__main__':
    a= []
    for _ in range(int(input())):
        name = input()
        score = float(input())
        a.append([score, name])

    a.sort()
    b = [i for i in a if i[0] != a[0][0]]
    c = [j for j in b if j[0] == b[0][0]]
    
    c.sort(key=lambda x: x[1])
    for i in range(len(c)):
        print(c[i][1])

Explanation

list a is simply a list of all the input data. It is sorted so that the lowest score is in the first index.

list b uses a list comprehension to remove all instances of the lowest score from list a. The lowest remaining score in the resulting list (found in the first index since it will be the first value taken from the already-sorted list a) is the second-lowest score overall.

list c then uses another list comprehension to trim down list b to only those scores that are equal to that second-lowest overall score.

list c is now a nested list containing the scores and names of the students that will make up the output. It is sorted alphabetically according to student name by using a lambda expression which indicates to use the contents of the second index of each object to be sorted ('name' in [score, name]) as the key for sorting.

Finally the names from list c are printed to satisfy the problem.

Easy pythonic code that anyone can understand

l = []
second_lowest_names = []
scores = set()

for _ in range(int(input())):
    name = input()
    score = float(input())
    l.append([name, score])
    scores.add(score)
        
second_lowest = sorted(scores)[1]

for name, score in l:
    if score == second_lowest:
        second_lowest_names.append(name)

for name in sorted(second_lowest_names):
    print(name, end='\n')

My shortest version, while still keeping things readable

n = int(raw_input())
marks = [[raw_input(), float(raw_input())] for i in  xrange(n)]           
scores = sorted({s[1] for s in marks})
result = sorted(s[0] for s in marks if s[1] == scores[1])
print '\n'.join(result)