I would like to complement the other very good answers.
As you know and as given by NASA's website, the definition of Total Impulse is
$$ I_T = \int_{0}^{\Delta t} T dt = \int_{0}^{\Delta t} \dot{m} v_e dt.$$
Which is, of course, the same equation as the impulse in basic physics.
If one considers the thrust as being almost constant over time, as is the case for many electric propulsion mission profiles and some very long burns of chemical rockets, the definition can be reduced to
$$ I_T = \Delta t \dot{m} v_e = m_p v_e = m_p I_{sp} g_0,$$
where $m_p$ is the total propellant mass. If you consider the Tsiolkovsky equation and substitute in it this new definition of specific impulse ($I_{sp} = I_T/m_p g_0$), we get
$$ \Delta V = \frac{I_T}{m_p} \ln \frac{m_0}{m_f} = \frac{I_T}{m_p} \ln \frac{m_f + m_p}{m_f} = \frac{I_T}{m_p} \ln\left( 1 + \frac{m_p}{m_f} \right) = \frac{I_T}{m_p} \left(\frac{m_p}{m_f} - \frac{1}{2} \left(\frac{m_p}{m_f}\right)^2 + \dots\right),$$
where the Taylor expansion of $\ln(1 + x)$ was used in the last term. If we consider now that $m_p \ll m_f$ (as in the case, again, of electric propulsion missions and other small propulsion systems), we can drop the higher order terms and get
$$ \Delta V \approx \frac{I_T}{m_f}.$$
Although this approximate relation might not be adequate for any precise estimate of a mission's $\Delta V$, it may serve to get some intuition about the physical meaning of the Total Impulse. We can observe, for example, that if we double the mass of the spacecraft we get half of the $\Delta V$ with the same total impulse.
More interestingly, if we write $I_T \approx m_f \Delta V$ it is possible to note that the total impulse holds a close relation to change of linear momentum, which was expected. This means that the total impulse gives the magnitude of momentum obtained by the vehicle after using its propulsion system.