As stated it is not true. Are there more conditions?
Let $ f(x,t) = \begin{cases} t^2 & |t| \leq 1 \\ t^4 & |t| \geq 1 \end{cases}$
Clearly $f$ satisfies (f1) as $f(x,u) = O(|u|^2) = o(|u|)$ as $u\to 0$ uniformly in $x$, and it satisfies $f(2)$ with $a_1 = a_2 = 1$, $s = 4$. For $N = 3$ we have that $(N+2)/(N-2) = 5 > 4$, so it is okay for $N = 1,2,3$.
Let $ u(x) = \frac{1}{(1 + |x|)^\frac14} $ clearly $u(x) \in L^5(\mathbb{R})$. But since $u \leq 1$, we have that
$ F(x,u) = \int_0^u f(x,t) \mathrm{d}t = \int_0^u t^2 \mathrm{d}t = \frac13 u^3 $
But the integral $\int_{\mathbb{R}} F(x,u) \mathrm{d}x$ does not converge, so $\Psi$ is not even $C(L^5(\mathbb{R}),\mathbb{R})$, never mind $C^1$.
Similarly for $N = 2$ we can take $u = (1+|x|)^{-1/2}$, and for $N = 3$ we can take $u = (1+|x|)^{-3/4}$ and produce analogous counterexamples.
The problem is that the decay condition $f(x,u) = o(|u|)$ is not strong enough for anything higher than $L^2$ control. Basically, to get the continuity (and similarly differentiability) of the functional, what you need to do is first cut off at a large radius $R$ (chosen relative to the function $u$ at which point you want to show continuity). Outside the large radius $R$ you have that $u$ and all functions $L^p$ close to $u$ have $L^p$ norm at most $\epsilon$, and must decay sufficiently fast. If we have a uniform in $x$ decay rate for $f(x,u)$ as $u\to 0$ that is sufficiently strong, then the integral of $F(x,u)$ outside $R$ is bounded by $L^p$ of $u$ outside $R$ and hence is negligible. Then we can work within the compact region where $u$ is large. Next you excise the points where $u$ blows up. By a Chebyshev inequality like argument these sets have small measure and neither $u$ nor $v$ can concentrate on them. So by excising some set on which $\int F(x,u)$ is of size $\epsilon$, we can assume $u$ is bounded. Now by uniform continuity of a continuous function on a compact set, we can conclude that $\int F(x,u)$ on this remaining set is continuous in $L^p$.
In short:
- Show that "non-compact" parts only contribute $\epsilon$
- Cut off the non-compact parts, the remaining part is compact and we can upgrade continuity of $f$ to uniform continuity.
I suspect (not having the CPDE paper in front of me at the moment) that (f1) should in fact be replaced by something with better decay properties. But as stated (f1) and (f2) are not sufficient to derive the desired conclusion.