To your first bullet point, you certainly could make it hyperbolic near the point. The (nonunique) geodesic connecting (-1,0) to (1,0) will go around the cusp and look semicircularish (but I don't think it will actually be a semicircle, just approximately one).
Perhaps something easier to visualize is that $\mathbb{R}^2 - \{pt\}$ is diffeomorphic to $S^1\times\mathbb{R}$. If you give this space the product metric of the usual metrics, then you can easily see and work out the details.
To you second bullet point, you should modify the statement slightly (and somewhat pedantically). Any proper nonempty open subset $U$ of a complete connected manifold $M$ is incomplete. To see this, let $p\in U$ and $q\in M-U$. By Hopf-Rinow, since $M$ is complete there is a geodesic $\gamma$ starting at $p$ and ending at $q$. Since $U$ is an open subset, it is totally geodesic: what $U$ thinks are geodesics are precisely what $M$ does. Thus $U$ thinks $\gamma$ is a geodesic which doesn't stay in $U$ for all time, hence $U$ is incomplete.
To the third bullet point, (you say "surface" then use $\mathbb{R}^n$), the theorem is true for $\mathbb{R}^n$ from Cheeger and Gromoll's Soul Theorem together with Perelman's proof of the Soul Conjecture. The soul theorem states that if $M$ is complete and has nonnegative sectional curvature, then $M$ has a compact totally convex totally geodesic submanifold $K$ (called the soul) so that $M$ is diffeomorphic to the normal bundle of $K$. The Soul conjecture asks: If $M$ has nonnegative curvature everywhere and a point with all sectional curvatures positive, must $K$ be a point?
Perelman proved the answer is yes: under these hypothesis, $K$ is a point. But a normal bundle of a point in a manifold is diffeomorphic to $\mathbb{R}^n$.
Finally, I have been told, but I have no idea about references/proofs/etc that every noncompact surface has some metric (necessarily incomplete if the surface isn't $\mathbb{R}^2$ by the above) of positive curvature. I'll try to dig up a reference tomorrow, if I remember to.
Suppose $\Gamma_1 = g \Gamma_2 g^{-1}$. Then $g:M \to M$ induces an isometry $\bar{g}:M/\Gamma_1 \to M/\Gamma_2$ since for every $\gamma_1 \in \Gamma_1,$ there is a (unique) $\gamma_2 \in \Gamma_2$ such that $g\gamma_1 m= \gamma_2 g m$.
On the other hand, suppose $\phi: M/\Gamma_1 \to M/\Gamma_2$ is an isometry. Since $M$ is simply connected and the actions are free and proper, $\Gamma_i$ identifies with the group of deck transformations of $\pi_i:M \to M/\Gamma_i$.
To work with fundamental groups we should really pick a basepoint. Choose $x_0 \in M/\Gamma_1$ and a lift $\tilde{x_0} \in M$. Also choose a lift $\tilde{y_0} \in M$ of $y_0=\phi(x_0) \in M/\Gamma_2$. Then there is a unique lift $\tilde{\phi}:M \to M$ such that $\tilde{\phi}(\tilde{x_0})=\tilde{y_0}$ fitting into your commutative diagram. Now let $\gamma_1 \in \Gamma_1$. Then $\tilde{\phi} \circ \gamma_1 \circ \tilde{\phi}^{-1}$ maps $\tilde{y_0}$ to another lift of $y_0$, and is therefore an element of $\Gamma_2$.
Best Answer
I believe that the local isometry condition tells you that your map has the sort of local covering property that is really at the heart of covering spaces. The problem is that the map may not be surjective. For instance, take the inclusion map from the punctured plane into the plane. This is clearly a local isometry, but it is not surjective. Thus the map does not have the covering property near the origin $(0,0)$ in the codomain.
So really, you should think of local isometries as being like covering spaces. After all, if you were just to complete the domain and codomain and uniquely extend the map $f$, you would have a covering space.