Is my current understanding of the fundamental of calculus correct?

Without limits, $\int f(x) dx$ just means antiderivative, so there is nothing to prove!

The FTC says if we DEFINE the function $F(x) \colon = \int_a^x f(t) dt$, where $f$ is continuous, THEN $F$ is differentiable and furthermore, $F'(x)=f(x)$.

The symbol $\int f(x) dx$ is defined as the antiderivative of $f$, so there is no theorem necessary. $\int f(x) dx$ is DEFINED as the (equivalence class of ) function such that $\frac{d}{dx}\int f(x) dx=f(x)$

A good way to understand the FTC is this: sum of of the change in the middle is the total at the edges. Let's unravel that. Think about $f$ as the amount of some fluid. Then $f^\prime$ is the change in the amount of fluid. Ask the question: what is the total amount that the fluid changed? To calculate this, you would just add the amount of change at each point. Integration is really an infinite summation, and so the total amount of change in the interval $[a,b]$ is

$$ TotalChange= \int_{a}^{b} f^\prime(t) dt $$

But how are things changing? If there is a total change in the amount of fluid, it has to be leaving through the boundaries. Things change going over the $b$ boundary by going positive, and change by going over the $a$ boundary by going in the negative direction, and so the other way of calculating the total change is

$$ TotalChange = f(b) - f(a) $$

In total, this means

$$ \int_{a}^{b} f^\prime(t) dt = f(b) - f(a), $$

which is the Fundamental Theorem of Calculus.

This generalizes. In Multivariable calculus you learn Green's Theorem and Stokes' Theorem. Then you learn Stokes' Theorem on manifolds. It's all the same idea: the sum of the change is equal to the sum along the boundary. We can write this succinctly as

$$ \int_{\Omega} d\omega = \int_{\partial \Omega} \omega $$

but it's really the idea that matters. Notice this is the same thing as before: the boundary of the interval $[a,b]$ is the points $a$ and $b$, so the sum of the change is the sum of the boundary. Mathematicians will say Stokes' theorem on manifolds is the true FTC, but it's all the same. If you understand this idea, then you understand the FTC.

Now for part two: why does that derivative formula work out? Let's just do it algebraically. Let $F$ be a function such that $F^\prime = f$. Then

$$ \int_{a}^{x} f(t)dt = f(x) - f(a) $$

by the FTC. But since $a$ is a constant, $f(a)$ is a constant, so taking the $x$ derivatives of both sides gives

$$ \frac{d}{dx} \int_{a}^{t} f(x)dx = f'(x). $$

What this is saying is that the antiderivative $\int$ is somewhat the same as integration from a constant $a$ to a variable $x$. The end is result may differ by some (arbitrary) constant, but when you take the derivative again, that constant drops out.

Note that I fluffed over all of the details about having $f^\prime$ continuous etc., once you know the idea you can work out the details.

While the common statements of Fundamental Theorem of Calculus normally assume that the function involved is continuous, this is not a necessary requirement. More generally we have

If $f$ is Riemann integrable over $[a, b]$ and $$F(x) = \int_{a}^{x}f(t)\,dt$$ then $F$ is continuous on $[a, b]$. Further if $f$ is continuous at some point $c \in [a, b]$ then $F$ is differentiable at $c$ and $F'(c) = f(c)$.

If $f$ is continuous on $[a, b]$ then it is Riemann integrable on $[a, b]$ and then $F'(x) = f(x)$ for all $x \in [a, b]$ where $$F(x) = \int_{a}^{x}f(t)\,dt$$ These two versions are normally called the First Fundamental Theorem of Calculus. There is a second FTC which goes as follows:

If $f$ is Riemann integrable on $[a, b]$ and if there exists a function $F$ such that $F'(x) = f(x)$ for all $x \in [a, b]$ then $$\int_{a}^{b}f(x)\,dx = F(b) - F(a)$$ Again note that if we assume $f$ to be continuous on $[a, b]$ then the first FTC (mentioned earlier) guarantees the existence of a function $F$ with $F'(x) = f(x)$ namely $$F(x) = \int_{a}^{x}f(t)\,dt$$ and then in that case the first and second FTC are in reality the same. The difference between first and second FTC comes into play only when the function $f$ under consideration is not continuous.

Further note that the function $F$ with $F'(x) = f(x)$ for all $x \in [a, b]$ is called an "anti-derivative" of $f$ and it is not unique for a given $f$. The second FTC tells us that if a function $f$ possesses an anti-derivative $F$ then its integral can be calculated as the difference between values taken by the anti-derivative at the end points of the interval of integration. This is in fact the most commonly used theorem in elementary calculus to evaluate integrals. In contrast the first FTC does not deal with calculation of integrals, but rather gives us an anti-derivative in concrete form under suitable circumstances.

The problem with your statements regarding FTC is that they are dealing with $f'$ instead of $f$. There is no specific need to consider the derivative $f'$. Also the way you have mentioned the theorems you should also give the conditions on $f'$ due to which these theorems are valid (because they are not valid for every derivative $f'$).

Also I can understand your frustration with the notation $$\int_{a}^{x}f(t)\,dt$$ and the use of two variables $x, t$. The confusion in this notation is caused because very rarely do the textbooks provide the definition of the symbol $$\int_{a}^{b}f(x)\,dx$$ properly. The notation $$\int_{a}^{b}f(x)\,dx$$ is used to denote the integral of a function $f$ over interval $[a, b]$ and hence is dependent on $f$ and $a$ and $b$. It has nothing to do with $x$ and the $x$ in $f(x)$ and $dx$ in the above notation is dummy. We could (and some books do) as well write $\int_{a}^{b}f$. By the same token the symbol $$\int_{a}^{x}f(t)\,dt$$ denotes a quantity which is dependent on $a$, $x$ and function $f$ and $t$ is dummy. Therefore we can consider this as a function of $x$ (letting $a$ to be constant) and denote it by $F(x)$ and think about its derivative $F'(x)$. The first FTC says that $F'(c) = f(c)$ when $f$ is continuous at $c$.

At the same time one could ask: why do we use the dummy variable in the integral notation? Well it is simply because the most convenient way to express a function $f$ is by giving a formula for $f(x)$ to calculate the value of $f$ at a generic point $x$. The use of $x$ in $dx$ part is because of a technical convenience while using substitution to evaluate integrals.