The definition of proper you quote is not the standard definition, and it is not strong enough to imply this result. Consider the following example: Let $G = S^1$ act on $M = \mathbb{R}^2\smallsetminus\{(0,0)\}$ in the usual way, i.e. by rotations about the origin. This action is clearly free. However, equip $G$ with the discrete topology. Note that this makes the topology on $G$ finer, so the action map $G\times M\to M$ is still continuous (the preimages of open sets were open using the usually topology on $G$, so by making the topology strictly finer, they are still open).
Now, I claim that this action satisfies the definition of proper you state in the first quote. Pick $x$, and then the orbit of $x$ is the circle about the origin of radius $\|x\|$. Pick $y$ not in the orbit of $x$ (so $\|x\|\neq\|y\|$), let $\epsilon = |\|x\| - \|y \||$, and let $V$ and $W$ be the $\epsilon/2$ balls about $x$ and $y$, respectively. It is easy to see that $gV\cap W = \emptyset$ for each $g \in G$, as $gV$ is contained in the "$\epsilon/2$-annulus" about the orbit of $x$, while $W$ doesn't intersect this annulus.
However, obviously, there is no open neighborhood $U$ of any $x$ such that the $gU$ are pairwise disjoint. In fact, for each $x\in M$, any open neighborhood $U$ of $x$ intersects the orbit $Gx$ at infinitely many points.
I'll also mention that the quoted definition of proper doesn't imply many of the other important consequences of proper actions. Proper actions of discrete groups (by the standard definition) have finite stabilizers. But consider the example of $\mathbb{Z}$ acting trivially on a point; it satisfies the quoted definition vacuously (only one orbit), but the stabilizer of the point is $\mathbb{Z}$. Similarly, proper actions by the usual definition have Hausdorff quotients (distinct orbits can be separated by open sets)--in my opinion, this is the the main reason for restricting to proper actions (at least for many purposes). However, the example here satisfies the quoted definition, but the orbits of $(1,0)$ and $(0,1)$ contain arbitrarily close points, as explained at that page.
I'm not familiar with the book you are using, but it sounds like they intended the standard definition: the map $G\times M \to M\times M$ given by $(g,x)\mapsto (x,gx)$ is a proper map, i.e. a closed map such that the preimage of each compact set is compact. tom Dieck's Transformation Groups gives a pretty nice treatment of this definition and its consequences in a general setting starting on page 27.
Orientations on a smooth $n$-manifold are equivalent to (equivalence classes of) nonvanishing $n$-forms, where $\omega \sim \eta$ if there's a positive smooth function $f$ with $f\omega = \eta$. Equivalently, because $C^\infty(M,\Bbb R^+)$ is connected, $\omega$ and $\eta$ are equivalent iff they are in the same path component of the space of nonvanishing $n$-forms.
Now pick an orientation $[\omega]$. If $f_t$ is a homotopy through diffeomorphisms between the identity and $f$, then $f_t^*\omega$ is a path in the space of $n$-forms between $\omega$ and $f^*\omega$; and hence, they're equivalent orientations.
Note that there is no good way of talking about whether or not a smooth map is orientation-preserving or reversing, because $d_pf$ needn't be an isomorphism - which is the condition you need to make sense of "$f$ is orientation-(blah) at $p$". This is why we restrict to diffeomorphisms here.
Best Answer
If the action is orientation-preserving, construct explicitly an orientation on the quotient. Fix an orientation on $M$. If $\bar x\in M/\Gamma$ is a point and $x\in M$ is such that $\pi(x)=\bar x$, the differential $d\pi_x:T_xM\to T_{\bar x}(M/\Gamma)$ is an isomorphism, so you can push the orientation of $T_xM$ given by the orientation of $M$ to one on $T_{\bar x}(M/\Gamma)$. Check that this depends only on $\bar x$ and not on the preimage $x$ chosen: this is where the hypothesis comes in. Finally, check that this way of orienting the tangent spaces to $M/\Gamma$ is in fact an orientation of $M/\Gamma$.
Can you do the converse?