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.