Are you sure about this?
I explained the logic in the code .
Anyways, doing what is asked is enough. Here is the solution in words.
The Question:
 A digit prime is a prime number whose sum of digits is also a prime. Find the number of digit primes in the given range (inclusive)
The Solution:

First, run sieve over 1 Million to filter prime numbers. Sieve of Eratosthenes or Sieve of Sundaram aim for the same → To find all primes less than or equal to a given integer N (in this case, 1 Million).

Now that we have all primes below 1 Million. We have to filter them further. To do that, we iterate over the primes, check if the sum of digits of i^{th} prime is also a Prime Number.
Example: 13 is a prime. But the sum of digits, viz., 1 + 3 = 4, is not a prime. So, 13 is not a Digit Prime
Another Example: 67 is a prime and sum of digits, viz., 6 + 7 = 13 is also a prime. So, 67 is a Digit Prime. 
Our task is to filter all such Primes below 10^6.

The 2nd loop does the same thing.
digitprime[i] = prime[i] & prime[sumOfDigits[i]];
This statement returns 1 if i is a Prime and the sum of digits of i is a prime as well. Otherwise, it returns 0. 
Now, digitprime[i] = 1 if i is a digit prime else 0.

In order to answer the queries efficiently, we make use of Prefix Sum.

Say, you have to count the number of i's in the range [a, b] where digitprime[i] is 1. To do that, prefix sum is used.

Prefix sum is an Array that stores the sum of values in a prefix of the Array.

Say our digitprime array is:
[0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, ...] that corresponds to
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
which means, 2, 3, 5, 7, 11, ... are digit primes. 
The Prefix Sum looks like:
[0, 0, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, \dots] for
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, \dots]
There are 4 digit primes below 10, 5 digit primes less than or equal to 11 etc.
Formally,prefix[i]
denotes the number of digit primes less than or equal to i. 
Using basic math, the number of Digit Primes in the range [a, b] is given by
prefix[b]  prefix[a  1]
.
This is the best I could explain.