Another way of phrasing this is to look at the transcendental lattices, which are the orthogonal complements of the Picard lattices.

One obtains in particular a morphism $T(A) \to T(Km(A))$ (by looking at the usual Kummer construction) which multiplies the intersection form by 2, but is an isomorphism as free abelian groups (i.e. they have the same rank). Since the rank of $T(A)$ is given by $6 - pic(A)$ and the rank of $T(Km(A))$ is given by $22 - pic(Km(A))$, the conclusion follows.

Another way of phrasing this is to look at the transcendental lattices, which are the orthogonal complements of the Picard lattices.

One obtains in particular a morphism $T(A) \to T(Km(A))$ (by looking at the usual Kummer construction) which multiplies the intersection form by 2, but is an isomorphism as free abelian groups (i.e. they have the same rank). Since the rank of $T(A)$ is given by $6 - pic(A)$ and the rank of $T(Km(A))$ is given by $22 - pic(Km(A))$, the conclusion follows.

Edit: The rank of $T(A)$ is $6 - pic(A)$, since $T(A)$ is the orthogonal complement of $Pic(A)$ in the lattice $H^2(A)$, which is of rank 6. Similarly, since $H^2(Km(A))$ is of rank 22, it follows that the rank of $T(Km(A))$ is $22 - pic(Km(A))$.