Quick answer: the tangent line may intersect the curve many times. For general curves, the tangent line at a point cannot be defined reasonably by counting the intersection points. Just think of the graph of $x \mapsto \sin x$ at $x=\pi/2$: the tangent line is the line $y=1$, which touches the graph infinitely many times.
What you say can be made rigorous for very particular curves, like conic sections.
On the other side, it all boils down to your definition of tangent line. There are situations where mathematicians prefer to allow a whole bunch of tangent lines, although there is no single straight line that can be pointed out as the tangent line. In elementary treatments, however, differentiability is somehow equivalent to the existence of a tangent line.
I agree that $f(x) = x^2$ has a lot of tangent lines. In fact, for each point on its graph, there is a tangent line. Let's plot a few of them
and then pick one out.
The one that is picked out is "the tangent to $f(x) = x^2$ at $x = 3/2$".
There are several ways to express an equation of a line. One way is point-slope form,
$$ y - y_0 = m(x - x_0) \text{,} $$
where $(x_0,y_0)$ is a point on the line and $m$ is the slope of the line. Point-slope form is convenient for tangent lines because we can use the function to get a point on the graph and the derivative to get the slope of the line tangent to that point of the graph.
For the one tangent line picked out above, $x = 3/2$. We can quickly compute \begin{align*}
f(3/2) &= (3/2)^2 = 9/4 \\
f'(3/2) &= 2(3/2) = 3 \text{.}
\end{align*}
So an equation of that line is
$$ y - 9/4 = 3(x - 3/2) \text{.} $$
(Rearranging to slope-intercept form, which may be more familiar, we have $y = 3x - 9/4$. The slope is $3$ any way we write it, but it is easy to see that the line meets the $y$-axis in the lower half-plane.)
So, what can we do? For any choice of $x$, we can use the function to get a point on the graph and use the derivative of that function to get the slope of the tangent line to the graph at that point. The first graph (reproduced here)
has the graph of $y = f(x)$ in blue. Pick one of the green lines: it meets the blue curve at one point and is parallel to the curve at that point. Each green line is a line, none of these are not lines. Two things change as we pick different points on the graph: the coordinates of the point and the slope of the line through that point. This means that the derivative has to be a function: for each point on the graph, we need a slope of the line tangent to that point. The function you obtained was $f'(x) = 2x$. And we can see this: for the green lines tangent to the left half of the curve, the slopes are negative because when $x$ is negative, so is $2x$. On the right half, the slopes are positive. Near $x =0$, $x$ is small, $2x$ is small, and the slopes are near zero (meaning close to horizontal). As we run off to the left or right (allowing $x$ to have large magnitude) the slopes become steeper in both directions.
So, yes, the derivative is a function -- it has to be able to tell you about different slopes of lines at different points on the graph of the function. If you draw all the tangent lines, you get a lot of tangent lines.
It is a coincidence that each tangent line in the above graphs meet the graph of $f$ exactly once. (Blah blah, convex region, blah blah... not what we're here to talk about.) A more general example is this: \begin{align*}
f(x) &= x^3 - x \\
f'(x) &= 3x^2 - 1
\end{align*}
picking $x = 1/2$, \begin{align*}
f(1/2) &= (1/2)^3 - (1/2) = -3/8 \\
f'(1/2) &= 3(1/2)^2 - 1 = -1/4
\end{align*}
So we expect to see a line through $(1/2,-3/8)$ with slope $-1/4$ that is tangent to the curve at that point.
It also happens that the line meets the curve somewhere else. This is not "important". The required properties are "intersects the graph at the point with the $x$-coordinate we picked, $1/2$" and "is tangent to the graph at that point". That this line also meets the graph again somewhere else is just what happens when you graph the line having those two properties.
Best Answer
No, the derivative $f'(a)$ is the exact slope of the line tangent to $(a, f(a))$. Note that by definition
$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
What the article is saying is that for very small $|h| > 0$, the slope of the secant line passing through $(a, f(a))$ and $(a + h, f(a + h))$ is a good approximation of $f'(a)$.