I'll try to be a succint as possible but answering your doubts as far as I can, but you made too many questions. Try to summarize as far as you can, adressing your main concerns that will clear the secondary ones.
A. I want to understand deep down intuitively the connection between the primitive function or antiderivative and the definite integral. I am already aware of at least this much:
The so-called indefinite integral is not an integral. Integrals can be represented as areas but the indefinite integral has no bounds so is not an area and therefore not an integral.
The indefinite integral, in my opinion, should be called "primitive" to avoid confusions, as many people call it. The idea is that we learn how to find derivatives, and then are told: "Well, but what about the inverse problem: If we have a known function, what function should we differentiate to get it?" And here comes the idea of primitive of a function, or rather, primitives. The primitive of a given function, which we denote by
$$\int{ f(x) }dx = F(x)$$
is a function such that $F'(x) = f(x) \text{ ; } (1)$.
The notation was used by Leibniz to denote a function that satisfied $(1)$, using the arbitrary constant $C$ to denote that the function wasn't unique - rather, there was a family of primitives of a given function $f$, since the derivative of a constant is null. This is simple notation, but it has nothing to do with $\int_a^b f$ in the sense that $\int_a^b f$ is a number representing the limit of an integral sum of $f$ over $I = (a,b)$, and $\displaystyle \int f(x) dx +C $ is representing a function. As you say, $$\int{ f(x) }dx + C = F(x)$$ is not an integral sum, but a symbolysm for the function that satisfies $(1)$.
B. The first part of the FTC, which is the thing that connects differentiation/antidifferentiation to the definite integral, is this: (...) etc.
Here you're getting confused. The FTC states:
Let $f$ be a continuous real-valued function defined on a closed interval $[a, b]$. Let $F$ be the function defined, for all $x$ in $[a, b]$, by
$$F(x) = \int_a^x f(t) dt $$
Then $$F'(x) = f(x)$$
for all $x$ in $[a, b]$.
This, in short, says that $F$ is indeed another primitive of $f$.
A consequence of this is the so called second FTC and a corollary. Since two primitives are only different by a constant, we should have
$$F(x) - \int_a^x f(t) dt = C$$
But then putting, $x = a$ gives
$$F(a) = C$$
which states that
$$F(x) - \int_a^x f(t) dt = F(a)$$
or
$$F(x) - F(a) = \int_a^x f(t) dt$$
Plugging in $b$ as the upper bound gives the famous corollary:
$$F(b) - F(a) = \int_a^b f(t) dt$$
which states that if $F$ is a primitive of $f$, the previous equality holds.
The second FTC says:
Let $f$ be a function defined on a closed interval $[a, b]$ that admits a primitive $F$ on $[a, b]$, i.e.:
$$f(x) = F'(x)$$
If $f$ is integrable on $[a, b]$ then
$$\int_a^b f(x) dx = F(b) - F(a)$$
(And this didn't depend on the continuity of $f$! You can try and plot the integrals dependeing on the upper bound of discountinous functions to see how the integral always "behaves" much better than the integrand.)
The connection between the primitives and the deinite integral is thus: If we know that a function admits a primitive in an interval, we can easily calculate it's definite integral by means of the primitive.
Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$\frac{d}{dx} \int f(x)dx = f(x)$$
$$\int \left( \frac{d}{dx} f(x) \right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $\int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$\frac{d}{dx} \int f(x)dx = \frac{d}{dx} \int x^2dx = \frac{d}{dx} \left( \frac{x^3}{3} + C \right) = \frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$\int \left( \frac{d}{dx} f(x) \right) dx = \int \left( \frac{d}{dx} x^2 \right) dx = \int 2xdx = \frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
Best Answer
No, not always.
In fact, the definite integral with variable bounds can be and is strictly more restrictive than the indefinite integral, in the most general case where both can be taken as sensibly existing to their fullest extent that we can say one defines the same kind of function as the other.
While it is true that if we have two functions $F_1$ and $F_2$ defined by definite integrals of the same function with different bounds, i.e.
$$F_1(x) := \int_{a_1}^{x} f(t)\ dt\ \mathrm{and}\ F_2(x) := \int_{a_2}^{x} f(t)\ dt$$
with $a_1 \ne a_2$, then we will have that $F_1 - F_2$ is a constant function, the range of possible such constants may be limited. A simple counterexample is to take $f(x) := \sin(x)$. Now the integral from any lower bound $a$ to $x$ is just $\int_a^x \sin(t)\ dt = \cos(a) - \cos(x)$, but here $\cos(a)$ - the "constant" of our integration - can only range in $[-1, 1]$ (and the difference of any two such "constants" only within $[-2, 2]$), but the constant in $\int \sin(x)\ dx = -\cos(x) + C$ can be any real number whatsoever, even one far outside this range, say assign $C := \mbox{Graham's number}$. Thus the integrals with variable bounds are not exhaustive of the full family of antiderivatives, which is what is represented by the indefinite integral, and if we are really pedantic should be truly written as the set it is: in "reality", not in the informalities of school calculus,
$$\int f(x)\ dx = \left\{ F(x) + C : C \in \mathbb{R} \right\}$$
where $F'(x) = f(x)$. Actually, if we want to be really correct - and in fact this is important but only in cases where the definite integral technically doesn't make sense as evaluable between arbitrary pairs of points - we should take
$$\int f(x)\ dx = \left\{ F(x) \in \mathbb{R}^\mathbb{R} : \mbox{$F'$ exists and $F'(x) = f(x)$}\ \right\}$$.