Let $G$, an algebraic group, act morphically on the affine variety $X$.
Then we can also have $G$ act on the affine algebra $K[X]$ as follows: $$\tau_x(f(y))=f(x^{-1}\cdot y),\qquad (x\in G, y\in X)$$
Then $\tau:G\to GL(K[X]),\quad \tau:x\mapsto \tau_x$.
Humphreys says that the reason that the inverse appears, is so that $\tau$ is a group homomorphism. But to me it seems that without the inverse it would be a group homomorphism, and with it, it isn't even a group homomorphism.
$\tau_{xy}(f(z))=f((xy)^{-1}\cdot z)=f(y^{-1}\cdot x^{-1}\cdot z)$ and $\tau_x\tau_yf(z)=\tau_xf(y^{-1}\cdot z)=f(x^{-1}\cdot y^{-1}\cdot z),$ so these seem to fail to be a homomorphism, where it is clear to see that without the inverse, it would be a group homomorphism.
What's the deal?
Here is another way to see this. For $x\in G$, let $m_x:X\rightarrow X$ be the map $y\mapsto x.y$. This is obviously an homomorphism : $m_{xy}=m_x\circ m_y$. Now, by definition the action on $K[X]$ is the following : $x.f:=f\circ m_{x^{-1}}$. Let us check that this is indeed an action : $$ (xy).f=f\circ {m_{(xy)^{-1}}}=f\circ m_{y^{-1}x^{-1}}=f\circ m_{y^{-1}}\circ m_{x^{-1}}=x.(f\circ m_{y^{-1}})=x.(y.f) $$
If you do the same reasoning with the action $x.f:=f\circ m_x$, you will see it fails to be a left action (it is in fact a right action).
Using a bit of the language of category theory, you have a left action of $G$ on $X$. Now if you consider $k[X]=\operatorname{Hom}_k(X,\mathbb{A}^1_k)$, then $G$ still acts on $k[X]$, but because it is contravariant in $X$ here, everything is turned around. So you get a right action on $k[X]$. Now this is the standard trick : using inverses, you can change every right action into a left action.