You would not "jump" to the end of the chain immediately. An epsilon transition is a convenience, a "way of modeling the systems whose current states are not precisely known". In fact, you would actually consider your self to be at all reachable (via epsilon transition) states at once. This is part of the non-determinism, you're not necessarily at one state, but rather can be at multiple at the same time.
In your example, starting at $q_0$, we have epsilon transitions allowing us to be in states: $\{q_0, q_1, q_2, q_4, q_5, q_6\}$ at the same time. This means all values that any of these states transition on, we can transition on. In your example, from $\{q_0, q_1, q_2, q_4, q_5, q_6\}$ we can transition on $\{1, 0\}$. In fact, because we're already at state $q_6$, we don't necessarily need to transition at all because we're already accepting.
For your second question, you would not be in an endless loop, similar to my previous explanation, having a loop of epsilon transitions would not affect how many states we're currently at. If we're at $q_5$, we're also at $q_6$, looping back to $q_5$ won't matter because we're already there. The importance of the loop is the fact that $q_5$ also transitions on $1$, so $q_6$ epsilon-transitioning to $q_5$ allows us to accept however many $1$'s we need.
I think an extremely helpful way to digest NFAs (because they can be confusing sometimes) is to apply the Powerset Construction or at least think about how it would be applied to your example. As fade2black said, the NFA will create many branches of computation, this construction will only create 1, which in my opinion makes it much easier to interpret.
In your example the powerset construction would look like this:
Set \ Transition | 1 | 0
-------------------------+--------------------------+-----
{q0, q1, q2, q4, q5, q6} | {q0, q1, q2, q4, q5, q6} | {q3}
{q3} | {q2, q4, q5, q6} | N/A
{q2, q4, q5, q6} | {q5, q6} | {q3}
{q5, q6} | {q5, q6} | N/A
We can then define:
$$\begin{align}
A & = \{q_0, q_1, q_2, q_4, q_5, q_6\}\\
B & = \{q_3\}\\
C & = \{q_2, q_4, q_5, q_6\}\\
D & = \{q_5, q_6\}\\
\end{align}$$
And get the resulting equivalent DFA:

Then running your input becomes trivial.