In showing $\pi_2(X, x_0) \cong \pi_2(X, x_1)$, why don't we need a reverse path? In the analogue proof for the fundamental group, we needed a path $\alpha$ from one base point $x_0$ to the other base point $x_1$ and then we needed the reverse path $\overline{\alpha}$ to complete the loop.

Why don't we need a reverse path for $\pi_n(X)$ when $n \ge 2$?
If we have a path from a point in $X$, say $x_0$, to a base point of the sphere, say $x_1$, it seems we should need a reverse path going from $x_1$ to $x_0$.

From Hatcher's book:

 A: We are using the "reverse path" (or really an entire family of "rotated" versions of the path, indexed by $\partial I^n$).  Think about what Hatcher's diagram would look like in the case $n=1$.  Instead of having a big square with a smaller square inside, you would have just a big interval with a smaller interval inside.  The complement of the smaller interval consists of two little intervals; on one of those little intervals you use $\alpha$, and on the other you use $\overline{\alpha}$.
On the other hand, in the case of the square, the complement of the smaller square is an entire family of "little intervals", one for each point of $\partial I^n$ (these are the diagonal lines drawn in the picture).  On each of these little intervals, you use $\alpha$, but these intervals are parametrized according to which point of $\partial I^n$ you're at: namely, you start $\alpha$ at the edge of the outer square, and end $\alpha$ when you reach the edge of the inner square.  So on the right side of the square, for instance, you are actually parametrizing these intervals from right to left instead of from left to right, which means you're using $\alpha$ "backwards".
