The answer is: $k = 2n - 3$. If all pairs of cities are directly connected (complete graph) then all roads from $A$ and all roads to $B$ must be oriented: $k \ge (n-1) + (n-1) - 1 = 2n - 3$ or if $\deg(A) = \deg(B) = n - 1$, all other vertices have degree $2$, then again we have to orient all $2(n-2) + 1 = 2n - 3$ roads. Now we prove that by orienting of at most $2n - 3$ roads the existence of a required road for any fixed road $L$ can be guaranteed. We say that a path containing not oriented roads $A_1A_2, A_2A_3, \dots, A_{l-1}A_l$ is minimal, if there is no road $A_lA_j$ with $1 \le i, j \le l, |i - j| \ne 1$. At the first stage we orient all roads belonging to the minimal path $\Gamma_1$ without self-intersections starting at $A$ and ending at $B$: $\Gamma_1 = \{A_1A_2, A_2A_3, \dots, A_{l-1}A_l\}$, where $A = A_1, A_l = B$. If $\Gamma_1$ includes all cities, then we are done. At the second stage we orient all roads of a minimal path $\Gamma_2$ starting at some city $B_1 = A_{p_1} \in \Gamma_1$ and ending at some city $B_m = A_{p_m} \in \Gamma_1$ and including new cities $B_2, B_3, \dots, B_{m-1}$ not lying on $\Gamma_1$. Such a path exists since our graph is 2-connected: there is a path between any two cities even after the elimination of any given road. We say that a path $B_1B_2, B_2B_3, \dots, B_{m-1}B_m$ is stretched, if there is no road $A_iB_j$ with $1 \le i \le p_1-1$ or $p_2+1 \le i \le l$ and $2 \le j \le m-1$. In other words, $\Gamma_2$ is stretched means that $B_1$ is closer to $A$ than any other city inside $\Gamma_2$ and $B_m$ is closer to $B$ than any other city inside $\Gamma_2$. If $\Gamma_1$ and $\Gamma_2$ contain all cities we are done. If not, at the third stage we orient all roads of some minimal and stretched path $\Gamma_3$, starting at some city on $\Gamma_1 \cup \Gamma_2$ and ending at some city lying on $\Gamma_1 \cup \Gamma_2$, including new cities. After $q$ steps $\Gamma = \cup_{i=1}^q \Gamma_i$ will contain all cities. Now any road $l$ either belongs to some $\Gamma_i$ or connects $\Gamma_i$ to $\Gamma_j$. If $l$ belongs to some $\Gamma_i$, it is already oriented and there is a path starting at $A$ passing through $l$ and ending at $B$. Indeed, let $l \in \Gamma_s$. This path has two parts, first part connecting $A$ with $l$ and the second part, connecting $l$ with $B$. In order to get the first part, we take an edge $l$ and go back through oriented edges $\Gamma_s$ till the first edge of $\Gamma_s$, then we go back through oriented edges of some other $\Gamma_i$ having intersection with $\Gamma_s$ ($i$ is not necessarily equal to $s-1$), and so on until getting $A$. In order to get the second part, we take an edge $l$ and go through oriented edges $\Gamma_s$ till the first edge of $\Gamma_s$, then we go through oriented edges of some other $\Gamma_i$ having intersection with $\Gamma_s$ ($i$ is not necessarily equal to $s+1$), and so on until getting $B$. This paths pass only through roads oriented at stages 1, 2, ..., $q$. Now suppose that $l$ does not belong to $\cup_{i=1}^q \Gamma_i$ and somehow oriented, say it connects a city $C \in \Gamma_i$ and $D \in \Gamma_j$. Then the required path again has two parts, first part connecting $A$ with $C$ and the second part connecting $D$ with $B$. We get these parts as in the first case. It can be readily checked that since constructed paths $\Gamma_1, \dots, \Gamma_q$ are minimal and stretched, the required path is correctly defined.
Now we note that the total number of roads oriented at stages 1, 2, ..., $q$ is at most $2n - 3$. Indeed, suppose that at some stage we draw a path $\Gamma_i$ including $l$ new cities. Then the number of oriented roads is $l + 1$. Thus, the total number of oriented roads is maximal if $\Gamma_1$ includes one road and each other $\Gamma_i$ included two roads as in the second example constructed above. In this case the total number of oriented roads is $2n - 3$. Done.