Let us denote the scary word by ~s~, and the favourite word by ~f~.

We use dynamic programming. State ~dp[i][j]~ means that we are on the ~i^\text{th}~ position in the scary word, and we want to write down the ~j^\text{th}~ letter of the favourite word. The transition is given by ~dp[i][j] = \min(dp[k][j+1]+c)~, where ~c~ is the time cost for the transition. The transition can be done for each pair of positions ~(i, k)~ such that ~s[i] = f[j]~, ~s[k] = f[j+1]~, and either ~s[k-1]~ or ~s[k+1]~ exists and is equal to ~f[j]~. We first move from position ~i~ to either position ~k-1~ or position ~k+1~, whichever is equal to ~f[j]~, using the second kind of move, and then to position ~k~ using the first kind of move (so we will write down ~f[j+1]~). The cost ~c~ is the sum of costs of the two moves. We need to be careful when both positions ~k-1~ and ~k+1~ contain ~f[j]~ and take the one that is closer.

The time complexity is ~\mathcal O(n^2 m)~. The problem can be solved in ~\mathcal O(nm)~ complexity, but that is left as an exercise to the reader.