I've got a continuous-jump process $(X_t)_{t\ge0}$ with generator $$(Af)(x)=\tilde\lambda(x)((\tilde\kappa f)(x)-f(x)),$$ where $$\tilde\lambda=\lambda+c$$ for some bounded measurable nonnegative $\lambda,c$ and $$\tilde\kappa f:=\frac\lambda{\tilde\lambda}\kappa f+\frac c{\tilde\lambda}\mu f,$$ where $\kappa,\mu$ are Markov kernel. I found the following code for the simulation:
Now, I know that if $(H_n)_{n\in\mathbb N_0}$ is a Markov chain with transition kernel $\tilde\kappa$ and $(Z_n)_{n\in\mathbb N_0}$ is an $\operatorname{Exp}(1)$-distributed process independent of $H$, then $$X_t=Y_n\Leftrightarrow t\in\left[\sum_{i=0}^{n-1}\frac{Z_i}{\tilde\lambda(Y_i)},\sum_{i=0}^n\frac{Z_i}{\tilde\lambda(Y_i)}\right)\tag1.$$
Is is clear to me that, given that $\tau_1$ and $\tau_2$ are independent, $$t_i-t_{i-1}=\min(\tau_1,\tau_2)\sim\operatorname{Exp}(\tilde\lambda(x_{i-1})).$$
However, simulating $\tilde\kappa$, should mean that we need to "sample with probability $\frac\lambda{\tilde\lambda}$ from $\kappa$ and with the other probability, i.e. $\frac c{\tilde\lambda}$, sample from $\mu$. When we want to implement this, we should need to draw a $u$ from $\mathcal U_{[0,\:1)}$ and if $u<\lambda$, sample $x_i$ from $\kappa(x_{i-1},\;\cdot\;)$ and else, i.e. if $u\ge\lambda$, then sample $x_i$ from $\mu(x_{i-1},\;\cdot\;)$.
However, they base this decision on whether $\tau_1<\tau_2$ or not. Is this somehow equivalent to what I described above? I don't get why this is correct.
I assume you meant $U[0, \tilde{\lambda})$ instead of $U[0, 1)$ in your question. Anyway, yes, it is, as $P(\tau_1 < \tau_2)$ is indeed just $\frac{\lambda}{\tilde{\lambda}}$. You can show this by conditioning
$$P(\tau_1 < \tau_2) = \int_0^\infty P(\tau_1 < \tau_2 \mid \tau_1 = t) \lambda e^{-\lambda t} dt = \frac{\lambda}{\tilde{\lambda}}$$
Also note that if you sample 2 independent exponentials, one with rate $\lambda$ and the other rate $c$, then their minimum has exponential distribution with rate $\tilde{\lambda}$. (This is Poisson merging.)