Let $K$ be a field. Let $\phi$, $\phi_1$ and $\phi_2$ be injective endomorphisms of $K$ satisfying $\phi=\phi_1\circ\phi_2=\phi_2\circ\phi_1$.
It is a classical result in Galois theory that $\phi$ (resp. $\phi_1$, $\phi_2$) can be extended non-uniquely to an endomorphism of $K^{\operatorname{sep}}$, which we denote by $\hat{\phi}$ (resp. $\hat{\phi}_1$, $\hat{\phi}_2$).
Since the extension is not unique, now we fix a choice of $\hat{\phi}$. The question is can we choose $\hat{\phi}_1$ and $\hat{\phi}_2$ simultaneously that satify $\hat{\phi}=\hat{\phi}_1\circ\hat{\phi}_2=\hat{\phi}_2\circ\hat{\phi}_1$?
There need not exist any extensions of $\phi_1$ and $\phi_2$ that commute. Indeed, such extensions need not even exist when talking about automorphisms of finite Galois field extensions. In terms of group theory, the question then becomes, if $G$ is a finite group with normal subgroup $H$ and you have two elements in $G/H$ that commute, can they be lifted to elements of $G$ that commute? The answer is no; for instance, if $G$ is the quaternion group $Q_8$ and $H=\{1,-1\}$, then the images of $i$ and $j$ in $G/H$ commute but they cannot be lifted to elements of $G$ that commute since $\pm i$ does not commute with $\pm j$ for any choices of signs.
Concretely, this gives a counterexample to your question as follows. Take any finite Galois extension $L/k$ with Galois group $Q_8$, and let $K\subset L$ be the fixed field of the subgroup $\{1,-1\}$. Then $Gal(K/k)\cong Q_8/\{1,-1\}$ has commuting elements $i$ and $j$ which cannot be extended to commuting elements of $Gal(L/k)$. These are then commuting endomorphisms of $K$ that cannot be extended to commuting endomorphisms of $K^{sep}$ (since such extensions would restrict to endomorphisms of $L$).