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
$$\langle \alpha, \; \;\beta p, \; \;\gamma p^2 \rangle \mapsto \langle \alpha, \; \;\beta, \; \;\gamma \rangle.$$



As far as a factor of $4,$ taking $\Delta = b^2 - 4 a c,$ we can descend when $\Delta \equiv 4 \pmod {16},$ as $x^2 +3y^2 \mapsto x^2 + xy+ y^2,$ and 
 $\Delta \equiv 0 \pmod {16},$ as $x^2 +20y^2 \mapsto x^2 +  5y^2.$ No descent for the others, for example $x^2 + y^2$ or $x^2 + 2 y^2.$