Suppose we have the Dirac "function" and consider it as a distribution. Then
\begin{equation} f(x)=\int f(x-t)\delta(t)dt \end{equation} for suitable functions on $\mathbb{R}^n$. Can we also do this with the dirac measure? and is there a relation between the two? This measure is defined as ${\displaystyle \delta _{x}(A)=\left\{{\begin{matrix}1,&{\textrm {if}}\ x\in A\\0,&{\textrm {if}}\ x\,\not \in \,A\end{matrix}}\right.}$ and one could expect that $\delta(t)dt$ and $\delta_{t}( \{ t \} )$ would be a similar thing. Is there a way to connect these things?
Notice that this is not the same as Dirac delta distribution and measure? where the $x$ in my notation is in the dirac "function". They should equal the same number.
On https://en.wikipedia.org/wiki/Dirac_measure we find the expression
$\int _{X}f(y)\,\mathrm {d} \delta _{x}(y)=f(x)$
what I am looking to have is the $x$ inside the $f$ instead but am not sure what kind of framework that is the best.
The way you write your question, it seems to me like you're more or less just caught up in notation. I think you'd be hardpressed to find an actual, mathematical difference between the Dirac $\delta$-distribution and the Dirac measure $\delta_0$.
Let me elaborate a bit. You define a distribution $\delta$ on $C_c^{\infty}(\mathbb{R}^d)$ by $$\langle \delta,f\rangle=f(0)$$ This is your defining feature of the Delta distribution. If you are more comfortable with notation that's indicative of integration you could write this as $$ \int f(x-t) \delta(t)\textrm{d}t=f(x), $$ but this is really just notation. The mathematically precise formulation would be the one above: $\delta$ is the functional that returns the value of $f$ at $0$.
Now, for any suitable measure $\mu$ on $\mathbb{R}^d$, we get the functional $$ \langle \mu,f\rangle=\int f\textrm{d}\mu $$ which, for $\delta_0$, the Dirac mass at $0$, gives you $$ \langle \delta_0,f\rangle=\int f\textrm{d}\delta_0=f(0) $$ so in the world of distributions, $\delta=\delta_0$. There is no difference.