My class has not been following a book and my professor's last bit of notation is a bit confusing to me.
This is the goal. We are given a path-connected space $Y$ and $H$ a subgroup of $\pi_1(Y,y)$. We want to find a cover $X$, of $Y$ such that $X$ is path-connected, $p(x)=y$, and the $im(p_*(\pi(X,x))=H$
So we ask what are the points of $X$ going to be
We look at all the points over $y$, the fibre $p^{-1}(y)$.
Since $X$ is path-connected, every point of $p^{-1}(y)$ can be connected by a path $\alpha$ starting at the given point $x$ and ending at $x'$
The $p \circ \alpha$ is a path in $Y$ starting and ending at $y$.
This says if $\textbf{this is where the notation confuses me}$
$[\gamma]=[p \circ \alpha] \in \pi_1(Y,y)$
$x' * [\gamma]=x$
How can $x' * [\gamma]=x$? Isn't this concatenation an entire path and not just one point??
That notation doesn't mean much to me, either. But to try to help:
I would guess that perhaps the prof has defined an action of $\pi_1(Y)$ on the fiber $p^{-1}(y)$, an action denoted by "*". So this says that the path $\gamma$, acting on $x'$, gives $x$. (The definition of the action would be "lift the path to one in X that starts at $x'$, and see where it ends; that endpoint is defined to be $x' * \gamma$"). [I make this guess because something like that is what people typically do in working with covering spaces.]
If you're looking for a book to peek at, you might want to check out "Algebraic Topology: An Introduction" by Massey, who does all this, and with more or less this notation.