I do not follow the notion of hyper-identities very clearly. In the last line in the snippet, how looks the substitution $$x_1x_2x_1$$ for $F$, what is the result and can I see that this is not satisfied in $B$.
I understand that we need to use $$F(x_1x_2x_1,x_1x_2x_1)\approx x_1x_2x_1$$
for all six terms like these:
$$t_1(x,y) = x, \; t_2(x,y) = y, \; t_3(x,y) = xy, \; t_4(x,y) = yx, \; t_5(x,y) = xyx, \; t_6(x,y) = yxy$$
but I do not know how to evaluate this and use it to conclude the non-satisfiability of this as a hyper-identity.
You called $t_5$ to the term the authors refer to as $x_1x_2x_1$, so what they are suggesting is that
$$t_5(x,t_5(y,z)) \approx t_5(t_5(x,y),z)$$
is not an identity satisfied by all bands. Notice that the identity evaluates to
$$xyzyx \approx xyxzxyx.$$
Now, it is not difficult to conclude that the above identity is not satisfied by all bands.
Indeed, the variety of bands is axiomatized by the associativity and the idempotency laws.
The associative law just enables us to get rid of parentheses, and associate the variables in whichever way we want, and we're already doing that; the idempotent law comes into play when we get two consecutive occurrences of the same variable (or the same term), which doesn't happen in the terms of the identity above.
Hence the terms cannot be further simplified and that law doesn't follow from the ones defining bands.
If you prefer to obtain a band not satisfying the identity, the reasoning is the same, and just consider $\mathbf F_{\mathcal B}(3)$, the free band on three generators.
Update. (About showing that $xyzyx\approx xyxzxyx$ is not valid in all bands.)
While the conclusion above is correct, I found meanwhile that the argument is wrong.
Indeed there are identities satisfied by the variety of bands and that don't follow from the above reasoning.
The ultimate result to derive these equations should be the one given in the following theorem. It seems like it was originally proved in [Green&Rees], but I couldn't get that paper, so I found it in [Gerhard]; other sources are certainly available.
A little notation before the result: if $t$ is a term, let us denote by $\bar{t}(0)$ the last variable appearing in $t$ (counting from left), and $t(0)$ the prefix of $t$ up to, and excluding, $\bar{t}(0)$. Define also $\bar{t}(1)$ to be the last variable appearing in $t$, but now counting from the right, and $t(1)$ is the suffix of $t$, counting from, and excluding $\bar{t}(1)$.
So, for example, in the term $f(x,y,z)=xyzyx$, we have $\bar{f}(0)=z$ and $f(0)=xy$; also $\bar{f}(1)=z$ and $f(1)=yx$
For the case of $f=xyzyx$ and $g=xyxzxyx$, we have that $$\bar{f}(0)=\bar{g}(0)=z=\bar{f}(1)=\bar{g}(1),$$ but $f(0)=xy$ and $g(0)=xyx$, and therefore, $f(0)\approx g(0)$ is not valid, for example, in a right-zero semigroup (where $xy=y$ and $xyx=x$); also $f(1)=yx$ and $g(1)=xyx$ and $f(1)\approx g(1)$ is not valid in a left-zero semigroup.
Some examples of identities satisfied by all bands that don't seem to follow by the reasoning I had previously presented, can be derived from the theorem: if $t_1(x,y,z)$ and $t_2(x,y,z)$ are terms in the variables $x,y,z$ and no one other, then $$xyzt_1(x,y,z)zyx \approx xyzt_2(x,y,z)zyx$$ is valid in all bands; in particular, with $t_1(x,y,z)=x$ and $t_2(x,y,z)=z$, we obtain $$xyzxzyx \approx xyzyx,$$ which is not obvious at all.
Another approach, and this is, I think, the one that is the easiest to follow, is to give an example of a band not satisfying the identity. Verifying the idempotency of a semigroup from its Cayley table is trivial, but we must also verify the associativity, to start with, and this can be a very exhaustive procedure. A better way is if it is a semigroup of transformation on a set $X$ (maps from $X$ to $X$); this way, we have associativity for granted and only need to show that the maps are idempotent and the set of chosen maps is closed under composition.
Let $X = \{a_0, a_1, a_2, a_3, a_4, a_5\}$, and the maps $f, g, k_3, k_4, k_5$, where $k_u:x\mapsto a_u$ is constant, and $f$ and $g$ are given by $$ \begin{array}{r|rr} &f &g\\ \hline a_0 &a_1 &a_2\\ a_1 &a_1 &a_1\\ a_2 &a_2 &a_2\\ a_3 &a_4 &a_5\\ a_4 &a_4 &a_4\\ a_5 &a_5 &a_5 \end{array} $$ It's easy to check that $\{f,g,k_3,k_4,k_5\}$ is closed under composition and that the semigroup $\langle \{f,g,k_3,k_4,k_5\}, \circ \rangle$ has the following Cayley table: $$ \begin{array}{c|ccccc} \circ &f &g &k_3 &k_4 &k_5\\ \hline f &f &g &k_4 &k_4 &k_5\\ g &f &g &k_5 &k_4 &k_5\\ k_3 &k_3 &k_3 &k_3 &k_3 &k_3\\ k_4 &k_4 &k_4 &k_4 &k_4 &k_4\\ k_5 &k_5 &k_5 &k_5 &k_5 &k_5 \end{array} $$ In this band we have $$fgk_3gf=fgk_3=fk_5=k_5,$$ but $$fgfk_3fgf=fgfk_3=fgk_4=fk_4=k_4,$$ so the identity $$xyzyx\approx xyxzxyx$$ doesn't hold.
[Gerhard] J.A. Gerhard, Equational classes of idempotent semigroups, Journal of Algebra, 15 (1970), 195—224.
[Green&Rees] J.A. Green and D. Rees, On semigroups in which $x^r=x$, Proc. Cambridge Phil. Soc., 48 (1952), 35—40