= Solution
By <continuity> of $g''$ at $\alpha$, for any $\epsilon>0$ choose $\delta(\epsilon)>0$ such that $|g''(s)|\le |g''(\alpha)|+\epsilon=:M$ whenever $|s-\alpha|<\delta(\epsilon)$. Twice using the <fundamental theorem of calculus> gives the remainder as
$$
g(t)-g(\alpha)-g'(\alpha)(t-\alpha)=\int_\alpha^t[g'(s)-g'(\alpha)]\,ds.
$$
The inner difference is $\int_\alpha^s g''(r)\,dr$ and has modulus at most $M|s-\alpha|$. Integrating this bound over the interval between $\alpha$ and $t$ proves, for either sign of $t-\alpha$,
$$
\boxed{|g(t)-g(\alpha)-g'(\alpha)(t-\alpha)|\le L|t-\alpha|^2,\qquad L=(|g''(\alpha)|+\epsilon)/2.}
$$
This is the local quadratic <Taylor remainder> estimate. The given $\epsilon$ may enter the bound on the second derivative; no stronger differentiability assumption is needed.
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