In the following composition is left to right and written exponentially, tensor products over $K$ are written $\otimes$.
The context is as follows: $A$ is a central-simple $K$-algebra, $L/K$ is finite Galois with group $G$ and $L$ a splitting field of $A$, $V$ is an $n$-dimensional $L$-vector space and $h:A\otimes L\to\operatorname{End}_L(V)$ is an isomorphism of $L$-algebras.
For every $\sigma\in G$ there is a $K$-automorphism $h^{-1}\sigma h$ of $\operatorname{End}_L(V)$, where $\sigma:A\otimes L\to A\otimes L$ is $a\otimes\lambda\mapsto a\otimes\lambda^\sigma$. Since $\operatorname{End}_L(V)$ is a simple finite-dimensional $K$-subalgebra of the $K$-CSA $\operatorname{End}_K(V)$, by Skolem-Noether there are units such that $x^{h^{-1}\sigma h}=u_\sigma^{-1}xu_\sigma$ for all $x\in\operatorname{End}_L(V)$ (if $B\subset A$ is a simple subalgebra of a CSA, then $\operatorname{Aut}_K(B)\cong N_A(B)/Z_A(B)^\times$, the quotient of the normaliser by the centraliser). Since $u_\sigma u_\tau$ and $u_{\sigma\tau}$ induce the same automorphism, there are $c_{\sigma,\tau}\in L^\times$ with $u_\sigma u_\tau=u_{\sigma\tau}c_{\sigma,\tau}$ (the map $G\times G\to L^\times,(\sigma,\tau)\mapsto c_{\sigma,\tau}$ is a $2$-cocycle). If we replace the $u_\sigma$ by $u_\sigma a_\sigma$ with $a_\sigma\in L^\times$, the inner automorphisms remain unchanged and $c_{\sigma,\tau}$ becomes $c_{\sigma,\tau}(a^\tau_\sigma a_\tau a_{\sigma\tau}^{-1})$ (the $2$-cocycle changes up to a $2$-coboundary).
Now for the question itself: the text claims that a change of the isomorphism $h$ or the $L$-vector space $V$ results in the same kind of change in the coefficients $c_{\sigma,\tau}$ (by a "simple application of Skolem-Noether", according to the text). I'm having trouble proving this, here's my attempt for the first statement
Let $g:A\otimes L\to\operatorname{End}_L(V)$ be another isomorphism of $L$-algebras with associated units $v_\sigma$, then $g^{-1}h$ is an automorphism of the $L$-CSA $\operatorname{End}_L(V)$, so by Skolem-Noether there is an $s\in \operatorname{End}_L(V)^\times$ with $x^{g^{-1}h}=s^{-1}xs$ and $x^{h^{-1}g}=s^{h^{-1}g}x(s^{h^{-1}g})^{-1}$. Then $$v_\sigma^{-1}xv_\sigma=x^{g^{-1}\sigma g}=x^{g^{-1}hh^{-1}\sigma hh^{-1}g}=(s^{-1}xs)^{h^{-1}\sigma hh^{-1}g}=(u_\sigma^{-1}s^{-1}xsu_\sigma)^{h^{-1}g}=s^{h^{-1}g}u^{-1}_\sigma s^{-1}xsu_\sigma(s^{h^{-1}g})^{-1}=(su_\sigma (s^{h^{-1}g})^{-1})^{-1}x(su_\sigma (s^{h^{-1}g})^{-1}),$$ so $v_\sigma$ and $su_\sigma (s^{h^{-1}g})^{-1}$ are related by a factor of $L^\times$.
How do I get rid of the ugly $s$-factors and make it so that $u_\sigma$ and $v_\sigma$ are related by a factor of $L^\times$? If $s$ were in $L^\times$, then $x^{g^{-1}h}=x$ and $g=h$, so there must be something else at play here. How do I prove the case of a different vector space?
This isn’t the symbolic proof you may be after, but it’s probably worth spelling out more coordinate independently. We have an $L$ algebra $B:=A\otimes L$ with certain properties (that can be seen using the isomorphism $h$, namely, being a matrix algebra over $L$).
One of these properties is that $B$ is a CSA over $K$, so in particular these Galois automorphisms (over $K$) are induced by inner automorphisms of $B$. We may lift these inner automorphisms to conjugation by elements of $B$, but we can’t necessarily do this preserving their composition. The obstruction to this is the $2$ cocycle valued in $L^*$.
Changing our choices of lifts gives us an equivalent $2$ cocycle, as you have checked, so the cohomology class is independent.
Note that to obtain this cocycle, and analyse its independence, we didn’t even need the choice of vector space $V$, we just needed the existence of such a $V$, to know that these Galois automorphisms (intrinsic to $B$) were induced by inner automorphisms.
This is all to say that in this instance, we can rephrase the argument more coordinate independently to see that we don’t need $V$ or $h$ at all, we just need the existence of such a $V$, $h$ to deduce properties about $B$. From this perspective, it’s clear that nothing is special about the choice of $V$ and $h$.