Electronic – LD1117 / SO-8 heat dissipation

For SMD there are adapter boards SMD-to-PTH, like

Soldering the parts takes some exercise. You may want parts with a 0.8mm pitch instead of the higher density 0.5mm pitch. The latter you'll find mostly on devices with 48 pins or more.
The other option is that you design your own boards for which you need an EDA (Elecronic Design Automation) package like Eagle. This has a free version with limitations, or a full version which has to be payed for. This design has a long learning curve. When you designed stuff this way you have to option of etching your own board or have it produced by a PCB production shop. The latter is the best choice for quality, the DIY is cheaper and faster.
For starters I recommend the adapter boards I mentioned at the beginning of this answer.

Good question.

Almost always, what damages things is heat, not current. The answer to your question is in the difference between local and global, or between sub-parts and the whole thing. The current may go through several sub-parts, and each one of them may have different power ratings. A bonding wire may have, by itself, a dissipation limit that is way below that of the whole device. A series sub-part, that needs to be there for whatever reason, may have a dissipation limit lower than that of the main sub-part. For instance, because it has a small volume, and a small surface in contact with the rest of the device, and cannot put out heat at the same rate as the main sub-part. In cases like these ones, it makes sense to specify a limit for the current, even after having specified a limit for the total dissipation, because those little, "secondary" sub-parts may not be able to dissipate heat at the same rate as the device as a whole.

Just an example. Imagine the situation in the following figure. The power rating for the whole device is 5 W. It would look like operating the device at \$V_d\$=2 V and \$I_d\$=1.5 A would be within safe limits, because \$P_d=V_d·I_d=2·1.5=3\$ W < 5 W. The global heat dissipation would be below its corresponding limit. However, \$I_d^2·0.1=1.5^2·0.1=0.225\$ W > 0.1 W, so that small sub-part would be exceeding its local dissipation limit, and hence the need to specify \$I_{dmax}\$ in addition to \$P_{max}\$. But the key is that, even for that current limit, the real reason behind it is heat, not current by itself. Except for a few cases (like electromigration), current by itself does not directly damage anything.