Either result can be obtained from the other. By adding a sufficiently large positive multiple of $I_n$ to $A$, we may always assume that $A$ is positive definite.
The answer by Joel Sjögren has indicated how to obtain the interlacing theorem for bordered matrices from the interlacing theorem for rank-one updates. More specifically, if we partition $A$ as $\pmatrix{A_{n-1}&b\\ b^\ast&c}$, then
$$
A^{-1}=\pmatrix{A_{n-1}^{-1}&0\\ 0&0}+\pmatrix{A_{n-1}^{-1}bs^{-1/2}\\ -s^{-1/2}}\pmatrix{s^{-1/2}b^\ast A_{n-1}^{-1}&-s^{-1/2}},
$$
where $s=c-b^\ast A_{n-1}^{-1}b$ is the Schur complement of $c$ in $A$. This is a rank-one update of $A_{n-1}^{-1}\oplus0$. Therefore the spectrum of $A_{n-1}^{-1}$ interlaces the spectrum of $A^{-1}$. In turn, the spectrum of $A_{n-1}$ interlaces the spectrum of $A$.
Conversely, Sylvester's secular theorem states that for any two rectangular matrices $X$ and $Y$, if $XY$ is a square matrix, $XY$ and $YX$ must have the same multi-set of non-zero eigenvalues. Therefore the eigenvalues of the positive definite matrix
$$
A+vv^\ast=\pmatrix{A^{1/2}&v}\pmatrix{A^{1/2}\\ v^\ast}
$$
are the $n$ positive eigenvalues of the $(n+1)\times(n+1)$ (singular) positive semidefinite matrix
$$
P:=\pmatrix{A^{1/2}\\ v^\ast}\pmatrix{A^{1/2}&v}=\pmatrix{A&A^{1/2}v\\ v^\ast A^{1/2}&v^\ast v}.
$$
Since $A$ is a bordered principal submatrix of $P$, its spectrum interlaces the spectrum of $P$ and the conclusion follows.
