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.
Your textbook needed an editor who could understand that the punctuation of the definition was rotten. It's be better to say
The point $x = c$ of a curve $y = f(x)$ defined by a function $f$ that's twice differentiable almost everywhere is an inflection point if either
(a) $f''(c) = 0$ and $f'$ changes sign as $x$ increases through $c$, or
(b) $f''(c)$ is undefined and $f'$ changes sign as $x$ increases through $c$.
The final clause ("The latter condition") is just plain wrong, as the example $y = x^4$ shows, because although $f''(0) = 0$, the concavity of the function is "up" on both sides of $x = 0$.
It also doesn't handle cases like $y = x^4 \sin (1/x)$ for $x \ne 0$ and $ y = 0$ for $x = 0$, which have second derivative zero, but for which the curvature changes sign infinitely often in any neighborhood of the origin -- one might call that an inflection point, or might not (I'd say "not", given the choice), but the authors' "as $x$ increases through $c$" suggests that they expect $f''$ to have one sign to the left of $c$ and the opposite sign to the right of $c$, at least locally, and for this function, that's just not true.
Best Answer
There is no contradiction. A graph of a function can have a tangent without being differentiable. Note that $x = 0$ (the $y$-axis) is a tangent to $f$ at $(0,0)$.