Explanation of trace function $\operatorname{Tr}_m(x) = x^{2^{m}} \oplus x$ The following statement is from a paper (Partitions in the S-Box of Streebog and
Kuznyechik) about S-Boxes:


  For all $ x \in \operatorname{GF}(2^{n})$, it holds that $x^{2^{n}} \oplus x = 0$.
  If $n= 2m$ then we define the trace from $\operatorname{GF}(2^{2m}) \to \operatorname{GF}(2^{m})$ as the function $\operatorname{Tr}_m(x) = x^{2^{m}} \oplus x$.


I don't understand why this works for all $ x \in \operatorname{GF}(2^{2m})$.
Why is it safe to say that this operation ends in the subfield $\operatorname{GF}(2^{m})$?
All elements of $\text{GF}(q)$ are roots of $x^q-x$. In fact, this is a litmus test for determining membership in $\text{GF}(q)$: when working in an extension field of $\text{GF}(q)$, say $\text{GF}(q^m)$, we can determine whether an $\alpha$ is a member of $\text{GF}(q)$ by computing $\alpha^q$ and checking whether the result equals $\alpha$ or not.

So, in $\text{GF}(2^n)$, $\alpha^{2^n} - \alpha = 0$, and if we remember that addition and subtraction are the same operation in fields of characteristic $2$ and that this operation is often denoted by $\oplus$, we have that $x^{2^n}\oplus x = 0$.

The trace function from $\text{GF}(q^k)$ to $\text{GF}(q)$ is defined as
$$\operatorname{Tr}(x) = x + x^q + x^{q^2} + \cdots + x^{q^{k-1}}.$$
Verify that for all $x \in \text{GF}(q^k)$, $\operatorname{Tr}(x)$ belongs to $\text{GF}(q)$.  (Hint: apply the litmus test). So for the special case when $k=2$, the trace function from  $\text{GF}(q^2)$ to $\text{GF}(q)$ is just $\operatorname{Tr}(x) = x + x^q$. I will leave it to the OP see what happens when $q$ equals $2^m$ and whether the statements of the S-Box book are true or not.