Since $a - b \equiv a + b \pmod 2,$ the running value after $k$ steps has the same parity as $a_1 + \cdots + a_k.$ So term $a_k$ is subtracted exactly when $a_k$ is even and the prefix sum $a_1 + \cdots + a_{k-1}$ is odd, and the final value is $12$ minus twice the total of the subtracted terms. We must count compositions of $12$ in which the even terms sitting where the prefix sum is odd total exactly $6.$ The prefix parity flips exactly at odd terms, so the odd terms come in $2m$ (the total is even), and the subtracted terms are precisely the even terms lying between the $(2i-1)$st and $2i$th odd terms; these $m$ "odd stretches" must hold even terms totaling $6,$ while the other $m + 1$ stretches hold even terms totaling $6 - A,$ where $A$ is the sum of the odd terms.

Let $f_r(t)$ be the number of ways to fill $r$ ordered stretches with sequences of even terms totaling $2t.$ One stretch is a composition of $2t$ into even parts, i.e. of $t:$ $f_1(t) = 2^{t-1}$ for $t \ge 1$ and $f_1(0) = 1;$ convolving gives the values needed below: $f_r(0) = 1,$ $f_r(1) = r,$ $f_2(2) = 5,$ and $f_1(3), f_2(3), f_3(3) = 4, 12, 25.$ Compositions of $A$ into $2m$ odd parts number $\binom{(A - 2m)/2 + 2m - 1}{2m - 1}.$

Casework on $m$ and $A:$ for $m = 1:$ $A = 2, 4, 6$ give $1 \cdot 4 \cdot f_2(2) = 20,$ $2 \cdot 4 \cdot f_2(1) = 16,$ and $3 \cdot 4 \cdot 1 = 12.$ For $m = 2:$ $A = 4, 6$ give $1 \cdot 12 \cdot f_3(1) = 36$ and $4 \cdot 12 \cdot 1 = 48.$ For $m = 3:$ $A = 6$ gives $1 \cdot 25 \cdot 1 = 25.$ The total is $20 + 16 + 12 + 36$ $+ 48 + 25 = 157.$