Concerning the noncompactness of the map, $I: W^{1, 2}(\mathbb{R}^n)\rightarrow L^{2^{\ast}}(\mathbb{R}^n)$ Consider the identity map $I:W^{1,2}(\mathbb{R^n})\rightarrow L^{2^{\ast}}(\mathbb{R}^n)$ where $n\geq 3$. Suppose that this map is not compact that is given some bounded sequence of functions $\{f_k\}_{k=1}^{\infty}\subset W^{1, 2}(\mathbb{R}^n)$ there is no subsequence that converges strongly to some $f$ with respect to the $L^{2^{\ast}}$ norm. So $f_k\nrightarrow f \text{ strongly in } L^{2^{\ast}}$.
So for some subsequence $\{f_{k_j}\}_{j=1}^{\infty}\subset \{f_k\}_{k=1}^{\infty}$ we can deduce that
\begin{equation}
f_{k_j}\rightharpoonup f \text{ in } L_{\text{loc}}^{2}(\mathbb{R}^n),\quad Df_{k_j}\rightharpoonup Df \text{ in } L^{2}(\mathbb{R}^n;\mathbb{R}^n)
\end{equation} 
since $W^{1, 2}(\mathbb{R}^n)$ is reflexive.
Is it possible to show that:


*

*$f_{k_j}\rightarrow f$ strongly in $L_{\text{loc}}^{2}(\mathbb{R}^n)$,

*$\vert Df_{k_j}\vert^2\rightharpoonup \mu\in\mathcal{M}(\mathbb{R}^n)$,

*$\vert f_{k_j}\vert^{2^{\ast}}\rightharpoonup \nu\in\mathcal{M}(\mathbb{R}^n)$


where $\mathcal{M}(\mathbb{R}^n)$ is the space of signed Radon measures on $\mathbb{R}^n$ with finite mass?
 A: Quite a few things here need straightening out. 

Suppose that this map is not compact 

This is not something we suppose. It is a  fact that $I$ is not compact.

given any bounded sequence of functions $\{f_k\}^\infty_{k=1}\subset W^{1,2}(\mathbb R^n)$ there is no subsequence that converges strongly

This is not what "not compact" means. In this context, "compact" means "for any bounded sequence there is a subsequence that converges". The negation of "compact" is "there exists a bounded sequence for which there is no subsequence that converges".  

We can deduce that

No, we cannot. We can deduce weak convergence for a subsequence of $f_k$, not for $f_k$ themselves.
Now to your questions: 


*

*$L^2_{\rm loc}$ is not a normed space, so   "strongly in $L^2_{\rm loc}$" is not a precise statement. Presumably you meant "strongly in $L^2(\Omega)$ for every bounded domain $\Omega$." Then it's true by Rellich-Kondrachov (since $2<2^*$) but not for the entire sequence - only for a subsequence.   

*and 3. The answer is yes - again, only for a subsequence - by virtue of weak* compactness of closed balls in the space of finite measures, considered as the dual of $C_0(\mathbb R^n)$. All you need from $f_k$ here is that $\int |Df_k|^2$ and $\int |f|^{2^*}$ are uniformly bounded; the first is true by assumption, the second by the Sobolev embedding. 
