If $\frac{1}{|f|}$ is integrable near $0$ then $\int_{0}^{t}\frac{1}{f}=\int_{0}^{1}\frac{\chi_{[0,t]}}{f}$, where $\chi_{[0,t]}$ is the indicator function of $[0,1]$. We have that $\frac{\chi_{[0,t]}}{f}\to0$ and $\left|\frac{\chi_{[0,t]}}{f}\right|\leq\left|\frac{1}{f}\right|$. We can apply dominated convergence theorem to get $$\lim_{t\to0}\int_{0}^{t}\frac{1}{f}=\lim_{t\to0}\int_{0}^{1}\frac{\chi_{[0,t]}}{f}=\int_{0}^{1}\lim_{t\to0}\frac{\chi_{[0,t]}}{f}=\int_{0}^{1}0=0$$
It remains to study the case in which $\frac{1}{f}$ is integrable but $\frac{1}{|f|}$ is not.
If we assume this is Henstock-Kurzweil integral, then we don't need $\frac{1}{|f|}$ to be integrable to apply dominated convergence. It is enough to know that $0\leq \frac{\chi_{[0,t]}}{f}\leq \frac{1}{f}$, and that $0$ and $\frac{1}{f}$ are integrable, which we have.
So, it remains to study the case when a less general integral definition is used, e.g. Riemann integral.
If it is Riemann integral ($\int_{0}^{t}\frac{1}{f}$ exists as an improper Riemann integral) then it is Henstock integrable and their values coincide.