Second derivative positive $\implies$ convex

You can prove this more or less my drawing pictures and looking at graphs. The way to make this intuition rigorous is usually by using the mean value theorem, I'll do it for you.

If $f$ is not convex it means that there is some interval $[a,b]$ where the line segment joining $f(a)$ and $f(b)$ is not always above the graph. Lets translate and rescale the argument of the function to make $a=0$, $b=1$. We are allowed to do this and it makes the algebra cleaner. Now there exists $c \in [0,1]$ such that $f(c)$ is above the line. By the mean value theorem there exists a point in $[0,c]$ where the derivative of $f$ is equal to the average rate of change between $f(0)$ and $f(c)$. But the average rate of change between $f(0)$ and $f(c)$ is greater than the slope of the line from $f(0)$ to f(1). Since $f'$ is increasing we have that after $c$, the function will be increasing too steeply to intersect the line from above at $f(1)$. This contradiction implies the function is convex.

To expand on Jonathan Y's comment, note that the argument of $f''$ in $(2.73)$ is $x^*$ and not $x_0$. The proof simply states that $x^*$ lies between $x$ and $x_0$. It turns, that you can pick such an $x^*$ such that $(2.73)$ is exact (that is, there a no higher order terms to begin with). Check out the Jonathan Y's link, look for the "Lagrange form" of the remainder, $R_k$, and plug in $k=1$.