Why can I have the positive and negative at the base of a transistor at the same time?
It's not a short circuit because the resistors restrict the current flow. 

You can't use just one pole of the battery, because there would not be a "circuit" if the current couldn't flow around. 

It isn't a transistor, that is a programmable unijunction transistor, and it has an anode, a cathode, and a gate (no base). 

By the way, contrary to what one might think, the 2N6027/6028 does not have just one junction, it has  three junctions (four layers). The name "programmable unijunction" comes from the fact it can be used in similar applications to an earlier device called a unijunction transistor. 

This is a simple LED blinker circuit ~ 1Hz. 

Edit: Even if you don't yet understand how the circuit works, you can see that the current is limited. Whatever G (gate) does, it can't be worse than shorted to ground (it actually does something like that when the PUT fires). Under those conditions the current through the 15K resistor is 6V/15K or about 0.4mA (not much current). The current through R3, no matter what happens with that A (anode) can't be any higher than 6V/470K or 13uA, which is quite tiny. 

Edit': The 15K and 27K resistors form a voltage divider. Before it fires, the PUT gate is pretty high impedance, we can ignore it for analysis. The voltage from the two resistors is 6V * 27K/(15K+27K) = 3.9V. The PUT will fire at about 0.35V more than 3.9V, or about 4.2V. 
RG (as shown on the datasheet) can be calculated as a bit under 10K.

So each flash, the capacitor starts off at a small voltage and charges to 4.2V through the 470K resistor, then discharges through the LED, and the cycle starts again. 

Why were those values chosen, instead of different values or leaving one resistor out entirely? 


You don't want to make the voltage divider voltage too high or the
capacitor  charging will not be consistent, or may not fire at all.
That means you need R2 and it has to be not too different from R1.

The voltage has to be high enough that the residual voltage on the
capacitor doesn't matter too much and you don't need a bigger
capacitor than necessary. For sure you need R1. 

There is a natural voltage to use that is about 2/3 the supply
voltage (1-1/e), which gives you a time that is about R multiplied
by C.
The parallel combination of R1 and R2, 10K in your example, must not be too high or the
PUT will latch on. It must not be too low or a lot of power will be wasted and it will be harder to trigger, perhaps stopping the oscillation. A good value to use is one that's on the datasheet because you don't have to guess about what it will do. There are two options- 1M (too high) and 10K (okay).