Do "$K/k$ twisted" representations exist?

In fact, for any field extension $K/k$ and any $k$-algebra $A$, if $M$ and $N$ are finite dimensional $A$-modules such that $M\otimes_kK\cong N\otimes_kK$ as $A\otimes_kK$-modules, then $M\cong N$ as $A$-modules. This is the Noether-Deuring Theorem (see, for example, (19.25) in Lam's "First Course in Noncommutative Rings"; he assumes $A$ is finite-dimensional, but that's not essential).

For finite field extensions the proof is very short. If $M\otimes_kK\cong N\otimes_kK$ as $A\otimes_kK$-modules, then, restricting to $A$, $M\otimes_kK\cong N\otimes_kK$ as $A$-modules. But if $[K:k]=n$, then as $A$-modules $M\otimes_kK\cong M^n$, the direct sum of $n$ copies of $M$, and $N\otimes_kK\cong N^n$. So $M^n\cong N^n$. Now just apply the Krull-Schmidt Theorem to deduce that$M\cong N$.