Suppose we have $\phi(x)\in C^\infty_c$ and: $$\delta_a\{\phi\}=\int \delta(x-a)\phi(x)dx=\phi(a)$$ $$\delta_a*\phi=\phi(x-a)$$ The convolution leads to a shifting the input.
Suppose we multiply the integral by $f(x)\in C^\infty$: $$\int \delta(x-a)\phi(x)f(x)dx=\phi(a)f(a)$$ I suppose the notation for this could be: $$\delta_a\{\phi f\}=f(a)\delta_a\{\phi\}=\int\delta(x-a)f(a)\phi(x)dx$$ But for the convolution is it: $$\delta_a*[\phi(x)f(x)] $$ or $$[f(a)\delta_a]*\phi(x) $$
I am pretty sure these expressions are not equivalent as the first shifts the product of the two while the second shifts $\phi$ but multiplies it by the value of $f(a)$. are both valid and just mean different things (ie we can choose whichever one we need)?
I think you confused yourself by writing $\delta_a\{\phi f\} = f(a)\delta_a\{\phi\}$. This is technically true... but I'd prefer it if you wrote this instead as $\delta_a\{\phi f\} = f(a)\phi(a)$.
The convolution of the distribution $\delta_a$ against $\phi(x)f(x)\in C^{\infty}_0$, should be notated as $\delta_a * [\phi f]$. This is because $$\delta_a * [\phi f] = \delta_{y-a}\{\phi(x-y)f(x-y)\} = \phi(x-a)f(x-a)$$ $$\overset{\text{if we abuse notation...}}{=}\int\delta(y-a)\phi(x-y)f(x-y)dy.$$ Note that the result $\phi(x-a)f(x-a)\in C^{\infty}_0$ as well.