$F[x]/(x^2)\cong F[x]/(x^2 - 1)$ if and only if F has characteristic 2

No, the problem is correct as stated: in particular those rings are not isomorphic in characteristic zero. I think your mistake is coming from reading too much into your description of the elements of both quotient rings as (being represented by) $f_0 + f_1 x$. This shows that they are isomorphic as $F$-vector spaces: i.e., they both have dimension $2$. But it doesn't show you that they are isomorphic as rings.

Hint: if the characteristic is not $2$, then $(x-1)$ and $(x+1)$ are comaximal ideals in the polynomial ring $F[x]$, so one can apply the Chinese Remainder Theorem to get a nice description of $F[x]/(x^2-1)$. In particular, you should find that the quotient is a reduced ring, i.e., has no nonzero nilpotent elements, unlike $F[x]/(x^2)$.

I guess you can see why the two rings are isomorphic in characteristic $2$?

Added: After you have solved this problem, it is enlightening to do a more general one: let $F$ be a field, $P(x) \in F[x]$ a polynomial, and try to give as explicit a description as you can of the quotient ring $F[x]/(P(x))$. For instance: when is it an integral domain? A field? A connected ring (i.e., with no idempotents other than $0$ and $1$)? A reduced ring (i.e., with no nilpotents other than $0$)? It turns out that everything depends upon the shape of the factorization of $P(x)$ into powers of distinct irreducible polynomials...

HINT $\ $ If $\rm\ char(F) \ne 2\ $ then $\rm\:F[x]/(x^2-1)\ \cong\ F[x]/(x-1) + F[x]/(x+1)\ \cong\ F^2\:$ has nontrivial idempotents, e.g. $\rm\:(0,1)\:,\:$ but $\rm\:F[x]/(x^2)\:$ does not (as is easily verified).