Your jump condition is wrong. To see this, let's assume for a moment that $\mu=1$ because it plays no real role in your formulation. Then, if you want to integrate the $\nabla p$ term and still want to get a symmetric formulation, you need to start with the first equation in the form
$$
k^{-1} \mathbf v + \nabla p = k^{-1} \mathbf f,
$$
which ultimately leads to the jump condition $[p]=0$ -- in other words, the pressure must be continuous.
How do you see that this is the right jump condition? Multiply the equation with a test function $\phi$ and integrate over a (arbitrary) part of the domain $\Omega_1$, for example one of the subdomains where $k$ is constant. After integrating by parts, you have
$$
(\phi,k^{-1} \mathbf v)_{\Omega_1} - (\nabla\cdot\phi, p)_{\Omega_1} + (\phi,p\mathbf n)_{\partial\Omega_1}= (\phi,k^{-1} \mathbf f)_{\Omega_1}.
$$
Now do the same on $\Omega_2=\Omega\backslash\Omega_1$ and you get
$$
(\phi,k^{-1} \mathbf v)_{\Omega_2} - (\nabla\cdot\phi, p)_{\Omega_2} + (\phi,p\mathbf n)_{\partial\Omega_2}= (\phi,k^{-1} \mathbf f)_{\Omega_2}.
$$
In the last equation, the sign of the normal vector is of course outward from $\Omega_2$, whereas in the first equation it is outward from $\Omega_1$. Now add these two equations together and you get
$$
(\phi,k^{-1} \mathbf v)_{\Omega} - (\nabla\cdot\phi, p)_{\Omega} + (\phi,p\mathbf n)_{\partial\Omega} + (\phi,[p]\mathbf n)_{\Gamma} = (\phi,k^{-1} \mathbf f)_{\Omega},
$$
where $\Gamma$ is the interface between $\Omega_1$ and $\Omega_2$ and $[p]$ is the jump of $p$ on the interface. $\mathbf n$ is the normal from $\Omega_1$ info $\Omega_2$.
We could, on the other hand, also have multiplied the original equation by $\phi$ and instead integrated over the entire domain right away. This would have shown that $\mathbf v$ must satisfy the equation
$$
(\phi,k^{-1} \mathbf v)_{\Omega} - (\nabla\cdot\phi, p)_{\Omega} + (\phi,p\mathbf n)_{\partial\Omega} = (\phi,k^{-1} \mathbf f)_{\Omega}.
$$
Comparing with the equation immediately above, it is clear that we have the condition
$$
(\phi \cdot \mathbf n,[p])_{\Gamma} = 0.
$$
Because the normal traces $\phi \cdot \mathbf n$ of functions in $H(div)$ are in $L^2(\Gamma)$, this implies that $[p]=0$ in $L^2(\Gamma)$.
