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The symbol may naturally be thought of as an element in the K-theory of X
appears in the nLab page on principal symbols for differential operators. What does this mean? Are they talking about K-theory or K-homology? How does one produce a class from the symbol of the operator?
It's a bit easier to see this using a slightly non-standard definition of topological K-theory. Given a locally compact Hausdorff space $X$, let $\bf{E}$ be a complex of vector bundles, i.e. a sequence
$$0 \to E_0 \xrightarrow{\alpha_0} E_1 \xrightarrow{\alpha_1} \ldots \xrightarrow{\alpha_{n-1}} E_n \to 0$$
where the $\alpha_i$'s are bundle maps and $\alpha_{i+1} \circ \alpha_i = 0$. The support of $\bf{E}$ is by definition the set of all $x \in X$ such that the fiber of $\bf{E}$ over $x$ is not exact. A homotopy between complexes $\bf{E}$ and $\bf{F}$ is a complex over $X \times [0,1]$ whose restriction to $X \times 0$ is isomorphic to $\bf{E}$ and whose restriction to $X \times 1$ is isomorphic to $\bf{F}$. Finally, declare that two compactly supported complexes of vector bundles over $X$ are equivalent if there is a compactly supported homotopy between them, and let $C(X)$ denote the set of all equivalence classes.
$C(X)$ is an abelian group under Whitney sum of complexes, and it has a subgroup $C_0(X)$ consisting of complexes with empty support, i.e. the complex is exact over every point in $X$.
Proposition: $K(X) \cong C(X) / C_0(X)$
There is a proof in Atiyah's book on K-theory, for example.
With that in hand, let $D$ be an elliptic operator mapping smooth sections of a vector bundle $E$ to smooth sections of a vector bundle $F$ over the same compact base manifold $M$. Let $\pi \colon TM \to M$ denote the tangent bundle of $M$. For each $x \in M$ and $V \in T_x M$, the symbol of $D$ is a linear map
$$\sigma(x, V) \colon E_x \to F_x$$
which varies smoothly in $x$ and $V$. Thus it defines a complex of vector bundles over TM:
$$0 \to \pi^* E \xrightarrow{\sigma} \pi^* F \to 0$$
The condition that $D$ is elliptic means that $\sigma$ is invertible - and hence the complex above is exact - outside of the zero section of $TM$, which is compact since $M$ is compact. So by the definition of K-theory above we get a class $[\sigma] \in K(TM)$.
