First, find the number of unordered subsets of $[n]$ with no difference of 1, of size $k$. The number of these is ${n - k + 1 \choose k}$. If we have a subset $a_1, a_2, \ldots, a_k$ of $[n]$ in increasing order, with none of the differences 1, then $a_1, a_2 - 1, a_3 - 2, \cdots, a_k - (k-1)$ is a subset of $[n-k+1]$ in increasing order, and this transformation can be reversed. For example, consider the subsets of $[7]$ of size 3 with no difference 1; these are
$$135, 136, 137, 146, 147, 157, 246, 247, 257, 357$$
(where I write $135$ for $\{1, 3, 5\}$ for conciseness). We can subtract 1 from the second element and 2 from the third element of each of these to get
$$123, 124, 125, 134, 135, 145, 234, 235, 245, 345$$
Both of these contain ${7 - 3 + 1 \choose 3} = 10$ elements.
Now, as you've already noticed, each unordered set of size $k$ gives rise to $k!$ ordered sets. For size $n$ we can have $k$ as large as $\lceil n/2 \rceil$. So we have
$$f(n) = \sum_{k=1}^{\lceil n/2 \rceil} {n-k+1 \choose k} k! = \sum_{k=1}^{\lceil n/2 \rceil} {(n-k+1)! \over (n-2k+1)!}$$
For numerical values, see https://oeis.org/A122852 (as has already been observed) and subtract one. For example,
\begin{align} f(7) &= \sum_{k=1}^4 {8-k \choose k} k! \\
&= {7 \choose 1} 1! + {6 \choose 2} 2! + {5 \choose 3} 3! + {4 \choose 4} 4! \\
&= 7 \times 1+ 15 \times 2 + 10 \times 6 + 1 \times 24 \\
&= 121 \end{align}
Best Answer
For the first part of the problem, we start with the observation that any unfriendly subset of $[n+2]$ either contains $n+2$ or does not contain $n+2$. We count the number of unfriendly subsets in each case separately.
We first count all unfriendly subsets of $[n+2]$ that contain $n+2$. We do this by establishing a bijection between unfriendly subsets of $[n]$ and those that we just described. Note that by adding $n+2$ to each of the unfriendly subsets of $[n]$, we get unfriendly subsets of $[n+2]$. Conversely, every unfriendly subset of $[n+2]$ certainly does not contain $n+1$ and by removing $n+2$ from each of them, we get unfriendly subsets of $[n]$. This means that the number of unfriendly subsets of $[n+2]$ that contain $n+2$ is equal to the number of unfriendly subsets of $[n]$ (which is equal to $U(n)$, of course).
Now we count unfriendly subsets of $[n+2]$ that do not contain $n+2$. It is clear that each of the sets falling into this category is also an unfriendly subset of $[n+1]$. Therefore, the number of unfriendly subsets of $[n+2]$ that do not contain $n+2$ is equal to the number of unfriendly subsets of $[n+1]$ (which is equal to $U(n+1)$).
Therefore, we arrive at $U(n+2)=U(n+1)+U(n)$ for each $n\in\mathbb{N}$.
Now, by a similar argument, we have $$U(n,k)=U(n-1,k)+U(n-2,k-1)$$ for valid $n$ and $k$.
It is clear that $U(n,0)=1$ since the only unfriendly $0$-subset of $[n]$ is the null set. Note that all singleton subsets of $[n]$ are unfriendly, and therefore $U(n,1)=n$.
The above recurrence relation for $U(n,k)$ can be rewritten as $$U(n-2, k-1)=U(n,k)-U(n-1,k)$$ Summing over $n$, we obtain $$\sum_{n=k+1}^m \: U(n-2,k-1) = \sum_{n=k+1}^m \:(U(n,k)-U(n-1,k))=U(m,k)-U(k,k)=U(m,k)$$ since $U(k,k)=0$ for all $k\geq 2$. Now, letting $k=2$ gives $$U(m,2)=\sum_{n=3}^m \: U(n-2,1)=\sum_{n=3}^m \: (n-2) = 1+2+\dots+(m-2)=\binom{m-1}{2}$$ Letting $k=3$ gives $$U(m,3)=\sum_{n=4}^m \: U(n-2,2)=\sum_{n=4}^m \: \binom{n-3}{2} =\binom{m-2}{3}$$ where we have used the infamous hockey-stick identity in the last step.
Now, it is easy to show (by induction on $k$) that $$U(m,k)=\binom{m-k+1}{k}$$