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.
I think the first FTC:
If $f: [a,b] \to \Bbb R$ is continuous then $F: [a,b] \to \Bbb R$ defined by $F(x)=\int_a^x f(t)dt$ is differentiable and $F'(x)=f(x)$ for all $x \in [a,b]$.
is what people mean by saying the integration (which defines $F$) is the inverse of differentiation (as we have found a function with derivative $f$).
The second FTC
If $f: [a,b] \to \Bbb R$ is Riemann-integrable on $[a,b]$ and we have a function $F: [a,b] \to \Bbb R$ such that $F'(x)=f(x)$ on $[a,b]$, then $\int_a^b f(x)dx=F(b)-F(a)$.
is more of a "recipe" to find an integral: the target is to compute the definite integral and the tool we're given is to find an antiderivative. So not an inverse as such but a method. It's a bit of an iffy one, as an antiderivative $F$ need not exist at all (except when $f$ is continuous and the first FTC gives us one, but not explicitly, but at least we know some solution exists, but we don't have it in computable form yet). I think the first is closer to giving a direct "inverse" connection between integration and differentiation (and is often used in other contexts when we differentiate wrt boundaries of integrals, etc.). But that's just one view.
The first FTC can be summarised as $$\frac{d}{dx}\int_a^x f(t)dt = f(x)$$ so "Applying the integration operator to $f$, followed by the differentiation operator gives us back $f$ again".
Best Answer
Basically, there's a type error: "$\int f(x)\,dx$" is a perfectly meaningful thing, but that thing is not a single function - rather, it's a set of functions.
The point is that a function doesn't have a unique antiderivative. For example, ${x^2\over 2}$ is an antiderivative of $x$ (with respect to $x$ of course), but so is ${x^2\over 2}-4217$. It's not the indefinite integral which is suspect, but rather the notation we use around it - specifically, the way we use "$=$." Properly speaking, $\int f(x)dx$ refers to a set of functions.
This is generally addressed by including a constant of integration, so that we write $$\int x\,dx={x^2\over 2}+C$$ to mean "The set of antiderivatives of $x$ is the set of functions of the form ${x^2\over 2} + C$ for $C\in\mathbb{R}$."