Page 74 in Bruce & Giblin's Curves and Singularities, 2nd ed. contains the following passage:
Let $f: I \to \mathbb R$ be smooth and define $\phi: I \times \mathbb R^2 \to \mathbb R$ by $\phi (t,u) = f(t) + u_1 t + u_2 t^2$. Then $\phi$ is a family of functions $\phi_u$ where $\phi_u (t) = \phi (t,u)$ and $\phi_0 = f$. Now let $g: I \to \mathbb R^2$ be $g(t) = (-f'(t) + t f''(t), - {1\over 2} f''(t))$. It is easy to show that $u \in im(g)$ if and only if there exists $t \in I$ where $\phi_u'(t) = \phi_u''(t) = 0$. Since $im(g)$ is the set of critical values of $g$, Sard's theorem says that for all $u$ outside a null set in $\mathbb R^2$, $\phi_u$ has no singularities other than of type $A_1$.
Note that a singularity $t$ of type $A_1$ is such that $\phi'_u(t) = 0$ and $\phi''_u(t) \neq 0$.
I think this is a mistake in the text since outside $im(g)$ we have either $\phi'_u(t) \neq 0$ or $\phi''_u(t) \neq 0$ or both.
So for a point $u \notin im (g)$ at which we have both $\phi'_u(t) \neq 0$ and $\phi''_u(t) \neq 0$ for all $t$ we don't seem to have an $A_1$ singularity.
What am I missing?
The points you're concerned about aren't singularities at all, as $t$ is a singularity only when $\phi_u'(t)=0$. The authors are saying that, for such $u$, if $\phi_u$ has a singular point, then it is of type $A_1$.