Why can't the second fundamental theorem of calculus be proved in just two lines?
The problem with your proof is the assertion
Now $dF$ is just the small change in $F$ and $f(x)dx$ represents the infinitesimal area bounded by the curve and the $x$ axis.
That is indeed intuitively clear, and is the essence of the idea behind the fundamental theorem of calculus. It's pretty much what Leibniz said. It may be obvious in retrospect, but it took Leibniz and Newton to realize it (though it was in the mathematical air at the time).
The problem calling that a "proof" is the use of the word "infinitesimal". Just what is an infinitesimal number? Without a formal definition, your proof isn't one.
It took mathematicians several centuries to straighten this out. One way to do that is the long proof with limits of Riemann sums you refer to. Another newer way is to make the idea of an infinitesimal number rigorous enough to justify your argument. That can be done, but it's not easy.
Edit in response to this new part of the question:
Plus we do this sort of thing in physics all the time.
Of course. We do it in mathematics too, because it can be turned into a rigorous argument if necessary. Knowing that, we don't have to write that argument every time, and can rely on our trained intuition. In fact you can safely use that intuition even if you don't personally know or understand how to formalize it.
Variations on your question come up a lot on this site. Here are some related questions and answers.
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Rigorous definition of "differential"
Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?
Allow me to translate your line "Multiplying both sides by $dx$, we obtain $dF=f(x)dx$." into what, interpreted strictly, you said:
"Pretending that the symbols $\mathrm{d}x$ and $\mathrm{d}F$ have existence outside of the symbol $\frac{\mathrm{d}F}{\mathrm{d}x}$, which is unjustified, we can multiply both sides by $\mathrm{d}x$, obtaining $$ 0 = \mathrm{d}F = f(x) \mathrm{d}x = 0 \text{,} $$ which, while true, has destroyed all information in our equation."
Why is this? Because $\frac{\mathrm{d}F}{\mathrm{d}x}$ is defined to be $$ \lim_{h \rightarrow 0} \frac{F(x+h) - F(x)}{(x+h) - x} \text{.} $$ Assuming this limit exists (which happily you have asserted), we could attempt to apply limit laws to obtain $$ \frac{\lim_{h \rightarrow 0} F(x+h) - F(x)}{\lim_{h \rightarrow 0} (x+h) - x} \text{.} $$ However, this gives a denominator of $0$, so is disallowed by the limit laws. (In fact, it gives $0/0$, suggesting that one should be more careful in explaining how one is sneaking up on this ratio.) Since you ignore this problem, you have multiplied both sides of your equation by $\mathrm{d}x = \lim_{h \rightarrow 0} (x+h) - x = 0$. Fortunately, your remaining left-hand-side is $\lim_{h \rightarrow 0} F(x+h) - F(x) = 0$. So you arrive at the true equation $0=0$, but this is completely uninformative. There are no infinitesimals (whatever those are) remaining.
In response to OP's general response:
An integral is a limit of sums of non-infinitesimal quantities. An integral cannot be the sum of infinitesimals because the sum of any number of zeroes, even infinitely many zeroes, is zero. This is quite easy to see by considering the (ordinal indexed) sequence of partial sums, which are always zero.
The derivative is an indeterminate form of type "$0/0$". The integral is an indeterminate form of the type "$\infty \cdot 0$". As I note above, we must be careful in how we sneak up on such forms to avoid absurdities.
Attempts to use infinitesimals rigorously failed. (From the "Continuity and Infinitesimals" article of the Stanford Encyclopedia of Philosophy)
However useful it may have been in practice, the concept of infinitesimal could scarcely withstand logical scrutiny. Derided by Berkeley in the 18th century as “ghosts of departed quantities”, in the 19th century execrated by Cantor as “cholera-bacilli” infecting mathematics, and in the 20th roundly condemned by Bertrand Russell as “unnecessary, erroneous, and self-contradictory”
You observe that it seems one must learn some other form of mathematics before attempting derivatives and integrals. I agree. To rigorously compute limits of difference quotients (derivatives) and limits of Riemann sums (integrals), one should first learn to find the limits of plain sequences. But there is a bootstrapping problem. As a consequence, in practice, we teach what one might call naive differentiation and integration in Calculus I/II/III and rigorous differentiation and integration in some class with a name like Advanced Calculus. The recipes for differentiating the common basket of functions (polynomials, trig function, exponentials, and logs) are simple enough to teach early on. But there is a full $\epsilon$-$\delta$ treatment of use to those who face functions not in that basket.
In the 20th century, there has been some progress in making infinitesimals rigorous. Useful articles are nonstandard analysis and dual numbers. (Aside: the first words of the nonstandard analysis article are
The history of calculus is fraught with philosophical debates about the meaning and logical validity of fluxions or infinitesimal numbers. The standard way to resolve these debates is to define the operations of calculus using epsilon–delta procedures rather than infinitesimals."
Since one wishes to perform mathematics starting from self-evident truths, one rejects objects with debatable meaning or questionable logical validity.) There are criticisms of nonstandard analysis. While I know that dual numbers can be used for automatic differentiation, I have never seen an attempt to use them as infinitesimals in a theory of integration.
Many answers here seem to suggest that what your argument lacks is merely a rigorous theory of infinitesimals.
No. Your argument is simply wrong, regardless of whether there is a clear meaning of infinitesimals. Note that your argument doesn't make use of the condition that $f$ is continuous (hence integrable). However, there are examples of $F$ whose derivatives $f$ are not integrable (see this thread for instance).