Let's look at an example: a damped mass on a spring, constrained to move left-and-right only, in outer space so there's no gravity. I'm going to assign a coordinate $x$ that indicates the displacement of the mass from its rest position. So $x = 0$ is the rest position.
This makes the configuration space $M$ be just $\mathbb R$; a tangent vector $v$ to $\mathbb R$ can be written as a pair $(p; u)$, where $p$ is the "basepoint" of the vector, and $u$ is the "tangent vector" part -- the thing that indicates velocity. Thus for the curve $\gamma(t) = t^2 + 1$, we have, since $gamma'(t) = 2t$, that the tangent vector to $\gamma$ at $t = 3$ is the vector
$(10; 6)$, where the $10$ indicates the position, and the $6$ is the "vector part". With this formulation (which works nicely because $M = \mathbb R$, we have that $TM \sim \mathbb R \times \mathbb R$, and $\pi: TM \to M : (p; u) \mapsto p$.
To any smooth path in $M$, there's an associated path in $TM$. Here's an example.
The path defined by $h(t) = t^2$ corresponds to the path
$$
H: \mathbb R \to TM : t \mapsto (t^2; 2t)
$$
in $TM$. I wrote $H$ simply by using $h(t)$ as the first coordinate, and $h'(t)$ as the second.
What about the other direction? If we have a path in $TM$, is there a path in $M$ that corresponds to it in the way just described?
Let's look at a particular path in $TM$, namely,
$$
s: \mathbb R \to TM : t \mapsto (t; 0),
$$
or, less formally, $s(t) = (t; 0)$.
That describes the zero-vector at every point of $\mathbb R$. Now suppose that $f$ is a path in $M$. We know that the corresponding path in $TM$ looks like
$$
F(t) = (f(t), f'(t))
$$
For this to be the path $s$, we'd need two things. For every $t$, we'd need
$$
f(t) = t \\
f'(t) = 0.
$$
Those two things are inconsistent, so there's no such path $f$.
What's all this have to do with $q$ and $\dot{q}$? In what I've written above, a typical point of $TM$ is $(p; u)$, where both $p$ and $u$ are real numbers. I've used the name $\pi$ (for "projection to the basepoint") for the function defined by $\pi(p;u) = p$, and I haven't given a name to the function that sends $(p; u)$ to just $u$.
A physicist gives rather different names to these two functions. S/he calls the first one $q$, and the second (alas), $\dot{q}$. Normally we give our coordinate single-letter names like $x$ and $y$, but ... well, after a while, you get used to this. For now, just think of $\dot{q}$ as a new letter in some bizarre alphabet.
For the function $H$ above, we have
$$
q \circ H (t) = t^2\\
\dot{q} \circ H (t) = 2t.
$$
For the function $s$, we have
$$
q \circ s (t) = t\\
\dot{q} \circ s (t) = 0.
$$
It gets worse. For a function like $H$ or $s$, it's conventional to entirely suppress the function name, and just say that for $H$, we have
$$
q(t) = t^2\\
\dot{q} (t) = 2t.
$$
while for $s$ we have
$$
q(t) = t\\
\dot{q} (t) = 0.
$$
In the first case, if you take the derivative of what's called $q(t)$, you find out that it's equal to $\dot{q}(t)$, making that suggestive notation a little more reasonable, maybe. For the second, that turns out not to be true.
To summarize so far: a path in the tangent space "corresponds" (in the sense demonstrated in the example of $h$ and $H$) to a path in the base space only if, in physicist notation, the function $t \mapsto \dot{q}(t)$ turns out to be the time-derivative of the function $t \mapsto q(t)$. THAT is what that cryptic thing in your text is trying to say.
What about the whole spring-and-mass thing? Well, so far I've said that certain paths in $TM$ are "nice" in the sense that they correspond to paths in $M$. Even so, not every nice path in $TM$ corresponds to something physically real. The claim of the Lagrangian formulation is that you can write down a function $L$ on $TM$ such that a path is "physically real" only if a certain expression involving $L$ and that path turns out to be zero. In the mass-spring system, those paths will turn out to be damped harmonic motion -- no surprise! But the main thing here is that the notions of the tangent space and "nice" paths are independent of the energy/Lagrangian of the situation, and are merely dependent on the constraints of the system. (Mine, for instance, made the motion one dimensional, so that the configuration space $M$ was just $\mathbb R$.)
Best Answer
The simplest reason for why we can do that is because
Given a function $f(x)$, if we can write it as $f(x,y)$ where $y = y(x)$, we can apply the identity $$ df = \frac {\partial{f}} {\partial{x}} dx + \frac {\partial{f}} {\partial{y}} dy$$
The derivation of this identity never makes the assumption that $x$ and $y$ have to be independent. The $only$ problem that can arise is that we might give $y$ values which are not allowed.
For example, if $y = x^2$, we find that $\frac {df} {dx}$ is the same as that calculated using the identity, but f(2,3) is not valid, as $y = 2^2 = 4$.
But when we talk about the Lagrangian, since at the end we use the identity that $\frac {dx} {dt} = v$, hence we are assured that we will never stumble upon any such problem.