Let $f:(0,1) \to \mathbb{R}$ be a $C^1$ map. Assume that $f|_{(0,1/2]}=0$, and that there exists a strictly decreasing sequence $t_n \to 1/2$, such that for every $n$, $f|_{(t_n-\epsilon_n,t_n+\epsilon_n)}$ is affine for some $\epsilon_n>0$.
Does $f|_{(0,1/2+\epsilon]}=0$ for some $\epsilon>0$?
Clearly, the slopes $m_n$ of the affine parts on $(t_n-\epsilon_n,t_n+\epsilon_n)$ converge to $f'(1/2)=0$.
I think it's not true. I'll define your function on $(-1,1)$ instead to make the formulas easier, where we'll converge to $0$ from the right. Assume that $f$ is constantly $0$ on $(-1,0]$.
Let $U_n$ be pairwise-disjoint, closed neighborhoods around the points $t_n$ respectively and let $f(x) = \frac{x}{2^n}$ on $U_n$. Then $f(t_n) \rightarrow 0$ and $f' \rightarrow 0$ as $n \rightarrow \infty$. Notice that if $x \in U_n$ then $f(x) < f(y)$ for $y \in U_{n+1}$ so there is a monotone, smooth extension of $f$ on the interval $[\min(U_n), \min(U_{n+1})]$.
To make sure it's $C^1$ at $0$ we need the derivative of $f$ to converge to $0$ between the intervals $U_n$. To ensure this, it's easiest to pick $t_n = 1/2^n$ and $U_n = [t_n - 1/8^n, t_n + 1/8^n]$ (you can do it for any $t_n$ but the intervals can't be getting too close to each other or you'll have to adjust the functions).
Then $$\min(U_n) - \max(U_{n+1}) = (1/2^n - 1/8^n) - (1/2^{n+1} + 1/8^{n+1}) = $$ $$= 1/2^{n+1} - 9/8^{n+1}$$
which is the distance between the two intervals, and $$f[\min(U_n)] - f[\max(U_{n+1})] = \frac{1/2^n - 1/8^n}{2^n} - \frac{1/2^{n+1} + 1/8^{n+1}}{2^{n+1}}=$$
$$ = \text{big mess} = \cdots = \frac{1}{2^{2n+2}}\Big[3-\frac{5 \cdot 2^{n+2}}{8^n} \Big] < \min(U_n) - \max(U_{n+1})$$
Especially by letting the smooth extension be a $1/2^n$-approximation of the straight line concatenation between the end points of $U_n$ and $U_{n+1}$, it will be $C^1$ at $0$. Here you need a theorem to the effect of "the derivative of the smooth approximation can take values between $f'|U_n$ and $f'|U_{n+1}$"; this is visually obvious but I don't remember the exact theorem to quote.