If $$ L \to L' = L +\frac{dF(q,t)}{dt}$$ the corresponding Hamiltonian becomes
$$ H \to H' = H - \frac{\partial F(q,t)}{\partial t} $$ as shown here. Moreover, the canonical momentum becomes $$ p \to P = p + \frac{\partial F}{\partial q} $$ while $$ q \to Q = q $$ as shown here.
These formulas allow us to check the invariance of Hamilton's equations explicitly. Concretely,
\begin{align}
\frac{dq}{dt} &= \frac{\partial H}{\partial p} \notag \\
\end{align}
becomes
\begin{align}
\frac{dQ}{dt} &= \frac{\partial H'}{\partial P} \\
\therefore \quad \frac{dq}{dt} &= \frac{\partial \left(H - \frac{\partial F(q,t)}{\partial t} \right)}{\partial P} \\
\therefore \quad \frac{dq}{dt} &= \frac{\partial H}{\partial P} - \frac{\partial }{\partial P} \left( \frac{\partial F(q,t)}{\partial t} \right) \\
\therefore \quad \frac{dq}{dt} &= \frac{\partial H}{\partial P} \\
\therefore \quad \frac{dq}{dt} &= \frac{\partial H}{\partial p} \frac{\partial P}{\partial p} \\
\therefore \quad \frac{dq}{dt} &= \frac{\partial H}{\partial p} \quad \checkmark
\end{align}
where I used that $F$ does not depend on $P$ and $$\frac{\partial P}{\partial p} =\frac{\partial }{\partial p} \left( p+ \frac{\partial F}{\partial q} \right) = 1. $$
Analogously, we can check Hamilton's second equation:
$$ \frac{dp}{dt}= -\frac{\partial H(q,p,t)}{\partial q} .$$
However, there is a subtlety. After the transformation, we have on the right-hand side $\frac{\partial H'(Q,P,t)}{\partial Q}$. But here we need take into account that $p$ also depends on $q$, since $ p \to P = p + \frac{\partial F(Q,t)}{\partial Q} $. Therefore
\begin{align}
\frac{\partial H'(Q,P,t)}{\partial Q} &= \frac{\partial H'(Q,p + \frac{\partial F}{\partial q} ,t)}{\partial Q} \\
&= \frac{\partial H'(Q,p,t)}{\partial Q} + \frac{\partial H(Q,p,t) }{\partial p} \frac{\partial p}{\partial Q} \\
&= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{\partial^2 F(Q,t)}{\partial Q \partial t} + \frac{\partial H(Q,p,t) }{\partial p} \frac{\partial p}{\partial Q} \\
&= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{\partial^2 F(Q,t)}{\partial Q \partial t} + \dot Q \frac{\partial \left(P- \frac{\partial F}{\partial q}
\right)}{\partial Q} \\
&= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{\partial^2 F(Q,t)}{\partial Q \partial t} - \dot Q \frac{\partial
}{\partial Q} \frac{\partial F}{\partial q} \\
&= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{d}{dt} \frac{\partial F}{\partial Q} \,.
\end{align}
where we used that
$$ \frac{d}{dt} \frac{\partial F}{\partial Q}= \frac{\partial^2 F(Q,t)}{\partial Q \partial t} + \dot Q \frac{\partial
}{\partial Q} \frac{\partial F}{\partial q} . $$
Using this, we can rewrite Hamilton's second equation after the transformation as follows:
\begin{align}
\frac{dP}{dt}&= -\frac{\partial H'(Q,P,t)}{\partial Q} \\
\therefore \quad \frac{d}{dt} \left( p+ \frac{\partial F(q,t)}{\partial q} \right) &= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{d}{dt} \frac{\partial F}{\partial Q} \\
\therefore \quad \frac{dp}{dt} + \frac{d}{dt} \left(\frac{\partial F(q,t)}{\partial q} \right)&= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{d}{dt} \frac{\partial F}{\partial Q} \\
\therefore \quad \frac{dp}{dt} &= -\frac{\partial H}{\partial q} \quad \checkmark
\end{align}
EDIT: The subtlety was also noted here, but unfortunately without an answer and a few years ago there was even a paper which didn't notice it and claimed that Hamilton's equations are not invariant.
Best Answer
The $m_{kl}$ is a "mass matrix". In Cartesian coordinates the kinetic energy is $$T = \sum_i \frac{m_i}{2} v_i^2$$ with $v_i^2 = v_{i,x}^2 + v_{i,y}^2 + v_{i,z}^2$. For simplicity let us consider the case of one particle. We can write this in matrix form as $$T = \frac{1}{2} (v_x, v_y, v_z) \begin{pmatrix} m_i & 0 & 0 \\ 0 & m_i & 0 \\ 0 & 0 & m_i \end{pmatrix} \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \frac{1}{2} \sum_{kl} m_{kl} v_k v_l.$$ The latter expression generalizes to the case of several particles, just imagine a bigger matrix.
But typically when using the Lagrangian formalism Cartesian coordinates are not the most convenient. Now, since $v_k = \dot{x_k}$, if we change coordinates to $\mathbf{q} = \mathbf{q}(\mathbf x)$, we have $$\dot{q}_k = \sum_k \frac{\partial q_k}{\partial x_j} \dot{x_j}$$ and since the coordinate change is invertible, $$\dot{x_k} = \sum_j \frac{\partial x_k}{\partial q_k} \dot{q}_j. $$ Therefore in the new coordinates the expression for the kinetic energy is $$T = \frac{1}{2} \sum_{ijkl} m_{kl} \frac{\partial x_k}{\partial q_i} \dot{q}_i \frac{\partial x_l}{\partial q_j}\dot{q}_j = \frac{1}{2} \tilde{m}_{kl} \dot{q}_k\dot{q}_l \tag{1} $$ where $\tilde{m}_{kl} = \frac{1}{2} \sum_{ij} m_{ij} \frac{\partial x_i}{\partial q_k} \frac{\partial x_j}{\partial q_l}.$ (This, of course, is the transformation law for a rank 2 tensor, because that is what the mass matrix is.)
Since we are allowed to do any coordinate transformations, the entries in the matrix $\tilde{m}_{kl}$ may be non-constant functions of the coordinates. Then we see that (1) is the first equation in your question.
For a concrete example where the mass matrix has non-constant entries you can try to find the mass matrix in spherical coordinates, then compare with the expression for the kinetic energy in spherical coordinates. (The latter should be in Goldstein if you don't remember it.)
For the second part of your question, it is really just Newton's first law. If the system is at rest in a position where $\partial V /\partial q_j = 0$, it is at rest with no forces acting on it. The more rigorous way to to understand it is to write down the Euler-Lagrange equations and check that $q_k(t) = q_k^0$ is a solution, then appeal to a uniqueness theorem for solutions to differential equations.