(The local regularity theorem) Let $P$ be a differential operator of order $k$ that is elliptic over $\bar{U}, U \subseteq \Bbb R^n$ relatively comapct. Let $k,l$ be integers, $f \in W^l$ and $u \in W^r$. Assume that $Pu=f$, this equation takes place in $W^{r-k}$. Then for each function $\mu \in C_c^\infty(U)$, $\mu u \in W^{l+k}$.
The spaces where $W^l$ is the completion of the Schwartz space $S(\Bbb R^n)$ under Sobolev $l$-norm. $$ ||f||_l^2 := \int (1+|\xi|^2)^l |\hat{f} (\xi)|^2 \, d\xi $$ For an open set $U \subseteq \Bbb R^n$, we define $W^l(U)$ to be the completion of $C_c^\infty(U)$ under the Sobolev $l$-norm.
What confuses me is the appearance of the integer $r$. Is $r$ supposed to equal to $l$?
Is the statement true even when $k=0$?
Original source maybe hepful: page 46, Theorem 3.5.1.
Depending on how you interpret the expression $Pu$, the prerequisite $u\in W^{r}(U)$ can be completely dispensed with. In fact there is the following stronger statement (concerning the regularising properties of an elliptic order $k$ differential operator $P$), which works for $u$ only being a distribution:
If $u \in \mathscr{D}'(U)$ and $Pu \in W^l(U)$, then $\mu u\in W^{l+k}(U)$ for all $\mu\in C_c^\infty(U)$.
Now I suppose that your lecture notes do not assume familiarity with distributions, so we need to make sense of $Pu$ in a more conservative way. Well, if $r\ge k$ and $u\in W^{r}(U)$, then you know what $Pu$ is and the statement reduces to the one you have mentioned.
To illustrate the difference between the two formulations, consider the following example: Let $$u\in L^1_{loc}(U) \text{ with }\int_U u \Delta \varphi = 0 \text{ for all } \varphi \in C_c^\infty(U). \tag{$\star$}$$ Now your formulation of elliptic regularity gives the following: If we additionally assume that $u \in W^2(U)$, then we can integrate by parts and see that $\Delta u = 0$, which lies in $W^l(U)$ for all $l\ge 0$. Since $\Delta$ is elliptic, this implies that $\mu u\in W^{l+2}(U)$ for all $l$ and thus $u$ is smooth. However, using the stronger formulation, we do not need to assume that $u$ has any weak derivative: In fact ($\star$) is enough to conclude that $u$ is smooth.
Concerning your second question: Yes, it works for $k=0$ and I would say that then the statement is trivial: An order zero $DO$ (with smooth coefficients) has the form $Pu = a u$, where $a\in C^\infty(U)$ and ellipticity means that $a(x)\neq 0$ for all $x\in U$. In particular $\mu u = (\mu a^{-1}) Pu$ and multiplying a function in $W^l(U)$ with $\mu a^{-1}\in C_c^\infty(U)$ certainly yields a function in $W^l(U)$ again.