There are usually two kind of complementary approximations which one use in band theory. One is assuming parabolic dispersion $$ E(k)=\frac{\hbar^2k^2}{2m_e^*} + E_0 $$ Neglecting the offset energy $E_0$, this is equivalent to the dispersion of a free electron (for example in the vacuum) which can be described (at low speed $<c$) by the Hamiltonian $$ H=\frac{p^2}{2m_e} $$ The first equation can be derived from the second by using $p=-i\hbar\partial_x$ and by assuming plane waves wacefunctions $\psi(r,x)\propto e^{i k x}$. In this sense, the mass $m_e^*$ is called effective, because it is in general different from the "real" mass of the electron. Now if you compare the first equation with your dispersion relation that you wrote: $$ E=\frac{\hbar^2k^2}{2m_e}\times \text{(something)}^{-1} + \text{(something else)} $$ you recognize that $$ m_e^*=m_e \times \text{(something)} $$ In general, you can simply define the effective mass in terms of the second derivative of the dispersion, that is $$ m_e^*=[(\partial_k^2 E)/\hbar^2]^{-1} $$ which again gives the same result.
Now regarding the second part of your question, I mentioned that there are two main kind of approximations in band theory. The second is the tight binding: $$ E=-2t\cos{k} $$ where I just consider the 1-dimensional case. Now close to $k\approx0$ one recovers the parabolic dispersion because one has $$ E=-2t\cos{k}\approx -2t + t k^2 $$ Therefore by comparing with the parabolic dispersion you can identify $$ t=\frac{\hbar^2}{2m_e^*} $$ which is the relation between the hopping parameter and the effective mass. Alternatively, if you want, you can again define the effective mass in terms of the 2nd derivative of the dispersions and obtain $$ m_e^*=[(\partial_k^2 E)/\hbar^2]^{-1}=\frac{\hbar^2}{2t} $$ Now, you can clearly see that the larger the hopping $t$, the smaller the effective mass, as the author says.
Long story short, the effective mass and the hopping are defined as $$ t=\frac{\hbar^2}{2m_e^*}=\frac12\partial_k^2 E(k) $$
Does it make sense?Edit:
To answer the last edit of the question, which has the equation $$E=\frac{\hbar^2\,k^2}{2\,m_e}\frac{m_e}{{m_e}^*}+t_0\frac{A}{a^2}\frac{m_e}{{m_e}^*}\implies E\propto \frac{t_0}{{m_e}^*}\implies {m_e}^*\stackrel{\eqref{*}}\propto t_0$$
I notice that these implications are false. The energy is proportional to the inverse of the effective mass. The energy is written as a sum of two terms. The first term is proportional to $k^2$ and can be written as $tk^2$ where $t=\hbar^2/(2m_e^*)$ is the hopping (as I wrote above). The second term do not depend on the momentum. The constant $t_0$ is not the hopping. Also, you cannot write that the energy is proportional to $t_0$ because you cannot just ignore the first term. Up to a constant term, the energy is proportional to $k^2$ and the proportional constant is $t\neq t_0$.