One also has the classical Bonnet-Myers and Synge theorems.
The Bonnet-Myers theorem states that if $M$ is complete with Ricci curvature bounded below $\delta > 0$, then $M$ is compact with finite fundamental group.
The idea of the proof is the opposite of the one in Matt E's answer. In positive (sectional) curvature, geodesics tend to get stuck together. One uses this to show that geodesics past a certain length (something like $\pi/\delta^2$) cannot continue to minimize.
It follows that if $B_r(0)\subseteq T_p M$ is any closed ball with radius $r > \pi/\delta^2$, then $exp(B_r(0)) = M$. In particular, you've written $M$ as the compact image of a continuous function so it's compact.
The theorem about the fundamental group is an immediate corollary: look at the universal cover $\tilde{M}$. One can "pull back" the metric so that the covering map is a local isometry. Hence, the same curvature estimates apply to $\tilde{M}$, so it, too, must be compact. But a compact manifold can only finitely cover something, so the fundamental group of $M$ is finite.
To push this to Ricci, one uses a nice trick: if $\text{Ric} > 0$, then the sectional curvature in some directions must be $> 0$, and one uses these directions only to show geodesics eventually stop minimizing.
The Synge theorems have a different style of proof. Here's one version of the theorem: suppose $M$ is compact with positive curvature, and $f:M\rightarrow M$ is an isometry. Suppose $\text{dim}(M)$ is even and $f$ is orientation preserving OR $\text{dim}(M)$ is odd and $f$ is orientation reversing. Then $f$ has a fixed point.
The proof is straightforward: suppose not. Choose $p\in M$ with $d(p,f(p))$ as small as possible. Choose a minimal geodesic from $p$ to $f(p)$. One then computes variations of the geodesic (this uses the parity of the dimension, if I recall correctly), and shows that some nearby geodesic is smaller, contradicting your choice of $\gamma$.
As a corollary, one learns that the fundamental group of an even dimension positively curved compact manifold is either trivial or $\mathbb{Z}/2\mathbb{Z}$. For, suppose $M$ satisfies all the hypothesis. Look at the deck group action on $\tilde{M}$. If it's orientation preserving, it has a fixed point, but the only deck transformation with a fixed point is the identity. If it's orientation reversing, then it's square is orientation preserving, and hence is the identity, so all elements are of order 2 in $\pi_1$. If there are 2 orientation reversing maps, then their product is orientation preserving and hence the identity, so there is at most one element of order 2.
Likewise, in odd dimensions, the corollary is that $M$ must be orientable, since the deck group must acts on $\tilde{M}$ by orientation preserving maps.
Finally, to just randomly answer one of your other questions, curvature can affect higher homotopy groups. As Matt E pointed out, in negative (really, nonpositive) curvature, the universal cover is $\mathbb{R}^n$, implying all higher homotopy groups vanish. In positive curvature, this is a very important and currently unsolved problem.