Solution:

Let $d_i = f(i) - f(i-1),$ so each $|d_i| \in \{1, 2, 3\}$ and $d_1 + \cdots + d_6 = 12.$ If $k$ of the differences were negative, the sum would be at most $3(6 - k) - k = 18 - 4k,$ so $k \le 1.$ If no difference is negative, the solutions of $a + b + c = 6,$ $a + 2b + 3c = 12$ (with $a, b, c$ counting $1$s, $2$s, $3$s) are $(0,6,0),$ $(1,4,1),$ $(2,2,2),$ $(3,0,3),$ and all such orderings satisfy every pair condition, giving $$\binom{6}{0,6,0} + \binom{6}{1,4,1} + \binom{6}{2,2,2} + \binom{6}{3,0,3} = 1 + 30 + 90 + 20 = 141.$$

If $d_1 \lt 0,$ then $|f(2)| \ge 2$ with $f(2) \le d_1 + 3$ forces $d_1 = -1$ and $d_2 = 3;$ the remaining four differences are positive and sum to $10,$ giving $\binom{4}{1,0,3} + \binom{4}{0,2,2} = 4 + 6 = 10$ functions, and the case $d_6 \lt 0$ is symmetric: $20$ in all. Finally, if $d_{n+1} \lt 0$ for some $n = 1, 2, 3, 4,$ the pair conditions $|f(n+1) - f(n-1)| \ge 2$ and $|f(n+2) - f(n)| \ge 2$ force $d_{n+1} = -1$ and $d_n = d_{n+2} = 3.$ The other three differences are positive and sum to $7,$ achievable as two $3$s and a $1$ or a $3$ and two $2$s, each in $3$ orders: $6$ ways for each of the $4$ positions, or $24$ functions.

The total is $141 + 20 + 24 = 185.$