Hint
Let $\lambda$ be an eigenvalue and $v$ be an eigenvector.
Estimate
$$\bar{v}^T(Q^TQ)v=(\bar{v}^T Q^T)(Qv)$$
Condition number depends primarily on singular values of $A$.
(which are the eigenvalues of $A^TA$)
Adding $cI$ can be described simply as a modification on eigenvalues.
In order to modify singular values of $A$, you need to add
$UcIV=cUV$, if $A=U D_{iag}(s_i) V$ is the singular value decomposition.
I think this reflects the difficulty of expressing simply the Condition number of $A+cI$.
(For real symmetric matrices $V=U^T$, for condition number of complex matrices, I guess you should use $A^*A$ instead of $A^TA$. Are the problem restricted to matrixes with real elements?)
Response to the edited question:
If $A=U D V$, where $D=D_{iag}(s_i)$, then
$A^TA=V^T D U^TU D V=V^TD^2V$.
If you add $cI$ to A, then
$(A+cI)=UDV+cI=U(D+cU^T V^T)V$, and
$(A+cI)^T(A+cI)=V^T (D^T+cVU)(D+cU^T V^T) V $
so the further $VU$ is from being diagonal, the further
your singular values might fall from $s_i+c$.
I am expecting that there is no exact formulation, but you can vizualise
the change of condition number numerically with random matrices, starting from symmetric ones, and adding small amount $\mu$ of non-symmetric part.
$A'=B+\mu C$, $B=1/2(A+A^T)$, $C=-1/2 (A-A^T)$.
Best Answer
In the book, the condition number refers to the matrix $A \in \mathbb{R}^{n \times n}$, not the function as you stated in the question.
The condition number of a matrix $A$ is defined as
$$ \kappa(A) = \|A\|_2\|A^{-1}\|_2,$$
where $\| \cdot \|_2$ is spectral norm of a matrix. It is known that the spectral norm of a matrix equals its maximum singular value
$$ \|A\|_2 = \sigma_{max}(A) $$
and that the maximum singular value of $A^{-1}$ equals 1 over the minimum singular value of $A$
$$ \sigma_{max}(A^{-1}) = 1 / \sigma_{min}(A).$$
Thus,
$$ \kappa(A) = \sigma_{max}(A) / \sigma_{min}(A).$$
If the matrix $A$ is normal (which means $A$ can be decomposed as $A=Q \Lambda Q^T$ where $Q$ is an orthogonal matrix and $\Lambda$ is a diagonal matrix whose entries are the eigenvalues of $A$), and using the fact that $\sigma_i(A) = \sqrt{\lambda_i(A^TA)}$, we have
$$ \sigma_{max}(A) = \sqrt{\lambda_{max}(A^TA)} = \sqrt{\lambda_{max}((Q\Lambda Q^T)^TQ\Lambda Q^T)} = \sqrt{\lambda_{max}(Q\Lambda^2Q^T)} = \sqrt{\lambda_{max}(A)^2} = |\lambda_{max}(A)|, $$
and then we have
$$ \kappa(A) = |\lambda_{max}(A)|/|\lambda_{min}(A)|.$$
However, in the book, I do not see where the author mentions that the matrix $A$ is normal.