Voltage Regulator – Using Heat Sink Via Holes Under LM2596

1. You're using a strange mix of curved, straight and 45° traces. The point is purely aesthetic for a design like this, but I would try to stick to 45° traces.
2. Yes, they're the solder pads.
3. Not if your PCB netlist matches the schematic's.
4. You'll have to choose actual parts first and check the drawings in the datasheet, or, if you have parts already, measure them. This goes especially for the capacitors, resistors are more forgiving about this (though you'll want to use as much as possible the same pitch for them).
5. If you don't want to place a header or connector there I would place them a bit more apart. You can use a 1-pin header as component.

Also try to be consistent with the symbols in your schematic. You mix American resistor symbols with a European potmeter symbol. (Personally, I don't like the American symbol at all; a schematic with lots of resistors IMO looks messy and frenetic, while the same with European symbols radiates rest & relaxation :-)).

edit
A comment on the double-sided board. This is a design that easily can be done on a single layer. It's worth practicing the layout part of your design. You'll learn how just swapping two components may simplify your routing significantly. For instance the red trace from R6 to the IC can be swapped to the other side by simply placing R6 between R4 and C4. Things like that. It will help you later on when you have more complex designs with hundreds of nets. And it can be fun, too. I like this kind of puzzles a lot. It's like playing planarity.
The problem with this double-sided design is that there are pads which are not accessible for soldering, because they're under a component on the component side, like with C3 or M1. If you don't have plated-through holes you can't be sure the wire is properly soldered to the pad. Worse, if the component's package sits on the pad you can almost be certain that there won't be a solder connection between pin and pad.

LED strips work at \$12\, \mathrm{V}\$, but the individual chips do not. They are connected in groups of three with one series resistor. Voltage drop of one LED is around \$3.6\, \mathrm{V}\$, so the current is \$\dfrac{0.24}{3.6}\cong 65\, \mathrm{mA}\$. For 30 blocks, it is \$30\cdot 65 \cong 2\, \mathrm{A}\$.

LM2596 is quite ancient, you could do better with something like TPS54340 or even better, with synchronous buck regulator likeLT8614. Newer chips have higher switching frequecies, which means that you can use smaller inductors with fatter wire and also smaller capacitors.

With only one block attached, it will work just fine (although less advanced regulators will have lower efficiency).

If you connect too much LEDs, you will overload and overheat the regulator and it will employ its overtemperature protection.

Edit: LEDs are grouped by 3 like this: