Golf a golf-scorer

Python 2, 72 70 bytes

f=lambda x,*T:[x*(-1)**T.count(x),~x%2*15,-x][(x-3)/8]+(T>()and f(*T))

At one point I wished Python treated 0**0 == 0 for once so that I could do (-condition)**num. Call like f(11, 10, 3, 1, 2, 2).

Previous 72-byte version:

f=lambda x,*T:[~x%2*15,x*(-1)**(x<3or T.count(x))][x<11]+(T>()and f(*T))

><>, 63 57 56+2 = 65 59 58 bytes

The input numbers are expected to be on the stack at program start, so +2 bytes for the -v flag. Try it online!

</!?lp6$+1g6:
3\0
?\::6g2%*{+}1+:b=
;\~16g-26g2*-c6gf*+n

As all unused values in the code field are initialised to 0, it can be used to work out how many of each value is present on the stack by getting the value at [value,6], incrementing it, and putting it back in the code field. The total is then calculated as:

T = 0 + {for x in 3 to 10, x*([x,6]%2)} - [1,6] - 2*[2,6] + 15*[12,6]

Edit: golfed 6 bytes off by restructuring the input and switching the calculation steps around. Previous version:

:6g1+$6pl0=?\
/-*2g62-g610/
c ;n$\
6:b=?/>::6g2%*{+}1+!
\gf*+3/

Edit 2: saved 1 byte, thanks to Sp3000

MATL, 27 26 bytes

3:10=s2\7M*G12=15*Gt3<*_vs

The input is a column array, that is, values are separated by semicolons.

Try it online! or verify all test cases (this adds a loop to take all inputs, and replaces G by 1$0G to push latest input).

Explanation

3:10=    % Take input implicitly. Compare with range [3 4 ... 10], with broadcast
s        % Sum of each column: how may threes, fours, ... tens there are
2\       % Modulo 2
7M       % Push [3 4 ... 10] again
*        % Element-wise multiply (the sum of this array is the score of 3...10)
G        % Push input again
12=      % Compare with 12, element-wise
15*      % Multiply by 15 (this is the score of 12)
G        % Push input again
t3<      % Duplicate. True for entries 1 or 2
*_       % Multiply and negate (the sum of this array is the score of 1, 2)
v        % Concatenate all stack concents into a vertical array
s        % Sum of array. Implicitly display