Let $m$ be the midpoint of $[a,b].$ We have $f(m)\le (f(a) + f(b))/2.$ This is the same as saying $(m,f(m))$ lies on or below the line through $(a,f(a)), (b,f(b)).$ Thus the slope from $(a,f(a))$ to $(m,f(m))$ is $\le$ the slope from $(m,f(m))$ to $(b,f(b)).$ Apply the MVT to see this implies $f'(c)\le f'(d)$ for some $c\in (a,m), d \in (m,b).$ Apply the MVT again: $ 0\le f'(d)-f'(c)= f''(z)(d-c)$ to see $f''(z)\ge 0.$
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
I think that one of main properties of strictly convex set is uniqueness of minimum (or maximum) of any linear functional on it.