One of the major geometric applications of the sort of analysis that you describe is to index theory for elliptic operators on manifolds. Using geometry one can often construct a differential operator (e.g. the Dolbeault operator on a complex manifold) whose Fredholm index is an interesting geometric invariant (e.g. the holomorphic Euler characteristic). The Atiyah-Singer index theorem computes the Fredholm index in terms of local geometric data, yielding an interesting result about the original geometry (e.g. the Hirzebruch-Riemann-Roch theorem).

To prove the Atiyah-Singer index theorem and to extend it in various directions, it is useful to replace the original differential operator with a bounded counterpart in a way which preserves the essential structure, such as the Fredholm index. This is mainly because bounded operators are easier to work with: you don't have to worry about subtle issues involving the domain of the operator and you have access to the tools of Hilbert space theory. The standard way to do this is to replace the differential operator $D$ with the pseudodifferential operator $\phi(D)$ where $\phi$ is a suitable bounded continuous function on the spectrum of $D$.

In an effort to make this a little less vague, let me give an example of a purely geometric statement whose proof uses some sophisticated pseudodifferential operator theory.

Theorem(Gromov-Lawson): Let $M$ be a compact manifold which admits a metric of nonpositive sectional curvature. Then $M$ admits no metric of positive scalar curvature.

A nontrivial corollary is that the $n$-torus admits no metric of positive scalar curvature. I would guess that the Cheeger-Gromov-Lawson paper that you linked to is aimed at nailing down some of the analysis involved in the proof of this theorem.

*Sketch of Wrong Proof:* The universal cover $\tilde{M}$ is diffeomorphic to $\mathbb{R}^n$ since $M$ has a metric of nonpositive sectional curvature, so $\tilde{M}$ admits a spin structure. Suppose $M$ has a metric of positive scalar curvature; lift this metric to $\tilde{M}$ and form the spinor Dirac operator $D$. The Fredholm index of this operator is nonzero, but according to the Lichnerowicz formula:
$$D^2 = \Delta + \frac{\kappa}{4}$$
where $\Delta$ is the Laplacian and $\kappa$ is the scalar curvature function. $\Delta$ is a positive operator, so $D$ has a gap around $0$ in its spectrum which implies that it is invertible and hence has Fredholm index $0$, a contradiction.

*What is Wrong:* Differential operators on non-compact manifolds are rarely Fredholm, so the part of the argument where we assert that the Fredholm index of the spinor Dirac operator on $\tilde{M}$ is nonzero makes no sense.

*How to Fix the Proof:* Let us change the notion of "index" so that it makes sense and has a shot at being nonzero for operators on non-compact manifolds. The "index" of an operator will be zero if the operator is invertible for any sensible notion of "index", so the proof above really will work. There are a number of different ways to adapt the notion of index to non-compact spaces, but most of them involve a specific analytical idea. Basic fact: a bounded operator on Hilbert space is Fredholm if and only if it is invertible modulo the ideal of compact operators. To get a notion of index which is more suitable for non-compact spaces, replace the ideal of compact operators with the algebra of operators which are compact when you compress them to a bounded subset of the non-compact space. But this poses a tricky analytical challenge: how can one construct bounded pseudodifferential operators which are invertible up to "compactly compressible" operators? Needless to say, lots of tricky estimates are involved...