Yes. It is about equivalence over $SL_2 \mathbb Z.$ Your form $f$ (primitively) represents some value not divisible by the fixed prime $p.$ Indeed, from original coefficients $\langle a,b,c \rangle$ we know that at least one of $a,c, a+b+c$ is not divisible by $p.$ We may therefore demand $a \neq 0 \pmod p$ in $\langle a,b,c \rangle.$
For an integer $t$ we have the equivalent form $$ \langle a, \; \;b + 2 a t, \; \;c + b t + a t^2 \rangle. $$ As $p$ is odd and $a \neq 0 \pmod p,$ there exists some $\delta$ such that $b + 2 a \delta \equiv 0 \pmod p.$
At this point, we now have $\langle a,b,c \rangle$ with $a \neq 0 \pmod p$ and $b \equiv 0 \pmod p.$ You have stated that $$ b^2 - 4 a c \equiv 0 \pmod {p^2}. $$ It follows that $c \equiv 0 \pmod {p^2}.$
We may now rewrite $$\langle \alpha, \; \;\beta p, \; \;\gamma p^2 \rangle.$$ The descent is $$\langle \alpha, \; \;\beta p, \; \;\gamma p^2 \rangle \mapsto \langle \alpha, \; \;\beta, \; \;\gamma \rangle.$$