Difficulties with reading "informal" and "non-rigorous" sections when studying Algebraic topology

Well, your example is sort of hard to read without a blackboard because a lot of different images are going on. First, it's also worth noting that the identification $D^4=D^2\times D^2$ is only true up to homeomorphism when working with the $2$-norm. Only for the $\infty$-norm is it an on-the-nose identity of spaces.

However, Hatcher wants to map to $\mathbb{R}^3$, and the canonical way of going there is via the stereographic projection from the north pole $\phi_N((x,y,z,w))=\frac{1}{1-w}( x,y,z)$ (when realising $S^3$ under the $2$-norm), which has inverse given by $(x,y,z)\mapsto \frac{2}{x^2+y^2+z^2+1}(x,y,z,0)+\frac{x^2+y^2+z^2-1}{x^2+y^2+z^2+1}(0,0,0,1)$. Note that $\phi_N^{-1}$ extends continuously to the $1$-point compactification by mapping $\infty$ to $(0,0,0,1)$.

Okay, so what's $S^1\times S^1$ in $S^3$ under the $2$-norm? Well, it's any set of the form $\{(x,y)\in(\mathbb{R}^2)^2| \|x\|_2=r,\|y\|_2=s\}$ with $r^2+s^2=1$. For simplicity, set $r=s=\frac{1}{\sqrt{2}}$. For such an element $(x_1,x_2,y_1,y_2),$ we see that, for fixed $y_2$, the image of $\phi(x,y)$ is two circles for $|y_2|\neq \frac{1}{\sqrt{2}}$ and a unique circle for $|y_2|=\frac{1}{\sqrt{2}}$ (since this forces $y_1=0$). Now, at the same time, we see that, as $y_2$ ranges over $[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}],$ the third component of the image ranges over an interval (by continuity). Hence, $\phi(S^1\times S^1)$ is, indeed, the usual torus shape. Hence, $\mathbb{R}^3\setminus \phi_N(S^1\times S^1)$ has two components. Let $U$ denote the "inside", i.e. the bounded component, the closure of which is homeomorphic to the solid torus $\partial D^2\times D^2$. Hence, so is $\phi^{-1}(\overline{U})$. However, consider the stereographic projection $\phi_S$ from the south pole instead, and you'll see that it completely flips the picture, i.e. $$ \phi_S(\phi_N^{-1}((\mathbb{R}^3\cup \{\infty\})\setminus U)=\overline{U} $$

Hence, the outside of the torus (with $\infty$), is homeomorphic to the inside of the torus, which exactly gives you Hatcher's claim.

As for generally understanding intuitive explanations, I really recommend trying to draw situations. Naturally, drawing in four dimensions is pretty hard, but you can actually make a drawing in $3D$, where you end up cutting the top and bottom of $S^2$ (which gives you a copy of $\{0,1\}\times D^2$), which leaves you with a middle part (homeomorphic to $[0,1]\times S^1$).

As a finishing anecdote, a wise, older student once told me: "There are two kinds of people who do algebraic topology: The ones that love all the diagrams, and try to ignore the geometric pictures, and the ones that love weird geometry and decide to live with all the diagrams." I think that sort of explains the weird interplay between the abstraction and the handwaving.

The answer of @WoolierThenThou is good for motivation and intuition. I want to explain something a bit different, namely how to read advanced mathematics books, and the hard work that is sometimes (often?) needed.

These books have heavy prerequisites, and you have to know those prerequisites. In the case of Hatcher's textbook, there are very heavy prerequisites in analytic geometry and topology. If you are not good in analytic geometry and/or topology, then you have to be ready to back up and learn what you need to know. What you wrote about your background gives me a clue: "I have not studied any 'topology' other than basic point set topology"; well, then you're going to need to study some topology. Picking up Hatcher knowing only some point set topology is something like picking up a book on Banach and Hilbert spaces knowing only some advanced calculus.

For example, let's take a phrase which comes up on the page preceding the one you are reading:

... the torus $S^1 \times S^1$ is embedded in $\mathbb R^3$ in the standard way.

The author is definitely assuming a strong prerequisite here: not just that you are familiar with $S^1 \times S^1$ and $\mathbb R^3$ as topological spaces, but that you also know standard embedding of $S^1 \times S^1$ in $\mathbb R^3$.

If you don't know that embedding, then yes, you're gonna be stuck, confused, etc. So the correct thing for you to do is put down that book and go figure out what embedding the author is referring to. It's an example in topology, so more elementary books on analytic geometry or topology might be a good place to start.

For instance, you might be able to recall an example from analytic geometry or multivariable calculus: a torus is often depicted as the surface of revolution obtained by revolving a certain circle in the $x,z$ plane around the $z$-axis in $x,y,z$ space, for instance the circle $(x-2)^2 + z^2 = 1$. And now you might be able to use that description, combined with your expertise in analytic geometry, to either write down a parametric formula $x=f(s,t)$, $y=g(s,t)$, $z=h(s,t)$ for the torus, or perhaps to write down an equation $F(x,y,z)=0$ whose solution set is that that torus (cylindrical coordinates help). And now you might be able to use, or develop, your expertise in elementary topology to convince yourself that this torus embedded in $\mathbb R^3$ that you've just specified is indeed homeomorphic to the product space $S^1 \times S^1$.

Now that you've got that settled, read on. Let's turn to something else from your question. You wrote in your post:

Although it is quite natural to regard $\partial D^2\times D^2$ as a solid torus, I cannot see any obvious reason why it can be identified with the solid torus bounded by the torus knot $K$.

Here a big part of the trouble looks like you simply misread the passage from Hatcher's book, which I'll quote:

The first solid torus $S^1 \times D^2$ can be identified with the compact region in $\mathbb R^3$ bounded by the standard torus $S^1 \times S^1$ containing $K$.

Notice, we are not talking about the "solid torus bounded by the torus knot $K$", instead we are talking about "the ...solid torus... in $\mathbb R^3$ ... bounded by the standard torus $S^1 \times S^1$..."

Well, assuming that you have just carefully verified the standard embedding of $S^1 \times S^1$ into $\mathbb R^3$ as just described, you should now have little trouble extending this to an embedding of the solid torus into $\mathbb R^3$: instead of simply revolving the circle $(x-2)^2 + z^2 = 1$, revolve the entire disc $(x-2)^2 + z^2 \le 1$.

There are plenty of other things you will encounter, one of which is nicely explained in the answer of @WoolierThanThou, namely the standard homeomorphism between $S^3$ and the one-point compactification of $\mathbb R^3$; if you don't know that homeomorphism, stop, put down Hatcher's book, and learn about it.

Once you know that homeomorphism, then you can continue on to "the second solid torus..."