Take for example $$
f(t) = \begin{cases}
-x^2 &\text{if $x < 0$} \\
x^2 &\text{if $x \geq 0$.}
\end{cases}
$$
For $x<0$ you have $f''(x) = -2$ while for $x > 0$ you have $f''(x) = 2$. $f$ is continuous as $0$, since $\lim_{t\to0^-} f(t) = \lim_{t\to0^+} f(t) = 0$, but since the second-order left-derivative $-2$ is different from the second-order right-derivative $2$ at zero, the second-order derivative doesn't exist there.
For your second question, maybe things are clearer if stated like this
If the second derivative is greater than zero or less than zero at some point $x$, that point cannot be an inflection point
This is quite reasonable - if the second derivative exists and is positive (negative) at some $x$, than the first derivative is continuous at $x$ and strictly increasing (decreasing) around $x$. In both cases, $x$ cannot be an inflection point, since at such a point the first derivative needs to have a local maximum or minimum.
But if the second derivative doesn't exist, then no such reasoning is possible, i.e. for such points you don't know anything about the possible behaviour of the first derivative.
Best Answer
Really $|x|$ is just a convenient shorthand for what's really a piecewise defined function. If you allow piecewise defined functions, then it's rather easy to come up with examples.
$$f(x)=x^2,x>0$$ $$f(x)=-x^2,x<0$$
does the trick quite nicely at $x=0$. To make a function that has no second derivative at finitely/countably(?) many arbitrary points you can simply shift and add multiple functions like these.