What Arnol'd is asking is about the behaviour of the solution on $\mathbb R \cup \{\infty\}$, which is obtained by "gluing together" the ends of the real line at a "point at infinity". You can also visualize this as an attempt to map $\mathbb R$ to a circle $S^1$ stereographically. Take a look at the image below. We draw a line from the North pole $N$ of the circle, through any point $P$ on the circle, and then we identify the point $P'$ on the real line with the point $P$ on the circle. You can try obtaining an explicit formula however it's pretty much clear that it's one-to-one and onto - except for the case when $P=N$ - that is, when $NP$ and the real line become parallel. If $P$ approaches $N$ in an anticlockwise direction, the corresponding $P'$ approaches $+\infty$. For the other direction, it approaches $-\infty$. The projective real line is simply defined as having that one extra point, making the stereographic projection a bijection.
Now we're coming to what Arnol'd is asking. If we stay on the real line, a solution to a differential equation can diverge in finite time. However on the projective real line, nothing "diverges", it just reaches the point at infinity in finite time. Or: it reaches the North pole in finite time. Now if we move $P$ to $N$ and then continue going, we will end up on the negative real line.
Arnol'd asks: what solutions are nonsingular on the projective line? Well, yes we want to avoid "singular points", but we can think of this as passing from the positive real axis to the negative real axis continuously, with just an added minus sign that indicates we are now on the negative real line.
The solution to the equation $\dot x = x^n,n>1$ is $$x(t)= c_n\left(t_{\infty}-t\right){}^{\frac{1}{1-n}},$$
where $c_n:= (n-1)^{\frac{1}{1-n}}$, and $t_{\infty}:=\frac{x(0)^{1-n}}{n-1}$, so that $x$ blows up at $t=t_{\infty}$.
This is already in a form of "series expansion" around $t_{\infty}$. For instance, we can see that for $n=2$, we have $$x(t)=c_2 \frac{1}{t_{\infty}-t},$$
so that $$\lim_{\epsilon\rightarrow 0} x(t_{\infty}-\epsilon)=+\infty$$ and $$\lim_{\epsilon\rightarrow 0} x(t_{\infty}+\epsilon)=-\infty.$$
This is good, this is what we were talking about. If you plot the solution, you can see that it blows up, reaching its vertical asymptote, and then reappearing to the right of the asymptote from below, from $x=-\infty$. However, for $n>2$,
$$\lim_{\epsilon\rightarrow 0} x(t_{\infty}-\epsilon)=\lim_{\epsilon\rightarrow 0} c_n \frac{1}{\epsilon^{\frac {1}{n-1}}},$$
yet
$$\lim_{\epsilon\rightarrow 0} x(t_{\infty}+\epsilon)=\lim_{\epsilon\rightarrow 0} c_n \frac{1}{(- \epsilon)^{\frac {1}{n-1}}}.$$
If we extract the minus, we get a factor of $(-1)^\frac {1}{1-n}=e^{\frac{i \pi}{1-n}}$, which is not a real number! Clearly we have gotten out of the projective real line by trying to extend time beyond $t_{\infty}$, and ended up on some new branch of the appropriate complex function. I hope this helped! And I will leave the asker to re-examine the cases $n=0,1$ from this point of view!
