Regarding the difficulties with the condition "$g$ is positive": it's true that no non-constant concave function can be positive on $\mathbb{R}^n$. It can be positive on its domain, but I argue that's not that useful: in an extended-real context, concave functions are defined to be $-\infty$ outside of their domain. However, you really don't need $g$ to be positive; you simply need to define that $f^2/g$ will have the domain $\mathop{\textrm{dom}}(f) \cap \{x\,|\,g(x)>0\}$. (You do, however, need $f$ to be nonnegative, for different reasons.)
Now to the original problem: proving that $f^2/g$ is convex. The easiest thing to do is take this in multiple steps. First, verify that $h(x,y)=x^2/y$, $\mathop{\textrm{dom}} h=\{(x,y)\,|\,y>0\}$ is convex. The Hessian test works well in this case.
Now convince yourself that $h_2(x,y)=\max\{x,0\}^2/y$, with the same domain as $f$, is also convex. For positive $x$, this coincides exactly with $h$; for negative $x$, it is identically zero; and it is continuous at $x=0$.
The reason we take this step is that we now have a function $h_2$ that is convex, non-decreasing in $x$, and non-increasing in $y$. Now we can use composition rules, such as found in Section 3.2 of Boyd & Vandenberghe. In particular, these rules confirm that $h_2(f(x),g(x))$ is convex if $f(x)$ is convex and $g(x)$ is concave.
Finally, note that if $f(x)$ is nonnegative, then $h_2(f(x),g(x))=h(f(x),g(x))$. Hence the original proposition is true.
Technically, we can skip the construction of $h_2$ if we are very careful in the application of the composition rules. The standard rule says that if $h,f$ are convex functions, and $h$ is nondecreasing, then $h(f(x))$ is convex. We can strengthen it to this: if $h,f$ are convex, and $h$ is nondecreasing on the range of $f$, then $h(f(x))$ is convex. With this modified composition rule, we can skip the construction of $h_2$.
Q1: The answer is no. If $g''(a)>0$ for some $a\in [0,1],$ then $g''>0$ in a neighborhood of $a$ by the continuity of $g''.$ Hence $g$ is strictly convex in that neighborhood. Similarly, if $g''(a)<0$ for some $a\in [0,1],$ then $g$ is strictly concave in that neighborhood. We're left with the case $f''\equiv 0.$ But this implies $f(x) = ax +b$ on $[0,1],$ hence $f$ is both convex and concave everywhere on $[0,1].$
Added later, in answer to the comment: It's actually possible for a $C^2$ function to have uncountably many inflection points. Suppose $K\subset [0,1]$ is uncountable, compact, and has no interior (the Cantor set is an example). Define
$$f(x)=\begin{cases}d(x,K)\sin (1/d(x,K)),&x\notin K\\ 0,& x\in K\end{cases}$$ Then $f$ is continuous, and $f$ takes on positive and negative values on any interval containing a point of $K.$ Define
$$g(x)=\int_0^x\int_0^t f(s)\,ds\,dt.$$
Then $g\in C^2[0,1]$ and $g''=f.$ It follows that every point of $K$ is an inflection point of $g.$
Best Answer
In the case $\frac{a+b}{2} \le x \le b$ we can use the concavity condition for $a < \frac{a+b}{2} < x$, which gives $$ 2 \ge f(\frac{a+b}{2}) \ge \frac{x-(a+b)/2}{x-a}f(a) + \frac{(a+b)/2-a}{x-a} f(x) \, . $$ Now use that $f(a) \ge 0$ and $x-a \le b-a$.
Graphically: Let $l$ be the line joining $(a, f(a))$ and $(\frac{a+b}{2}, f(\frac{a+b}{2}))$. Then $f(x) \le l(x) \le 4$ for $\frac{a+b}{2} \le x \le b$.
The case $a \le x \le \frac{a+b}{2}$ works similarly, or can be reduced to the first case because of the symmetry of the problem.