How to interpret step-up output current with respect to input/output voltage
There is a fair bit of assumption in this as the "datasheet" is not very explicit. You would be much better off if we had a schematic of the actual circuit (you would be able to look up the converter datasheet and as a bonus you would get an insight into designing your own boost converter).

However, I would assume from their graph that they don't rate the U3V12F12 beyond 280mA - so I would not use it for your application. Although there are many, many alternatives (including designing your own board) I am assuming you would rather stick within the Polulu range, and your alternative part seems to do exactly what you want. The graph shows it is still characterised up to 700mA output - so we're happy. But how much current will it pull?

Let's examine the case at 300mA output - which you specified as worst case.

\$Power_{out}=Power_{in}\times\eta\$, where \$\eta\$ is the efficiency of the converter. Looking at the graph, we can see that \$\eta\approx0.88\$ at 300mA out. We also know that \$Power = Voltage \times Current = V \times I\$.

So \$V_{out}\times I_{out} = V_{in} \times I_{in} \times \eta\$ and hence \$12\times 0.3 = 3.7 \times I_{in} \times 0.88\$. 

From this, \$I_{in} = \frac{12 \times 0.3}{3.7 \times 0.88}= 1.1A\$ 

So, for the second converter choice, 300mA (0.3A in my calculations) requires around 1.1A in.

As for LC current spikes - these only occur when reactive loads (such as the L (inductor) and C (capacitor) in the name) cause voltage spikes due to decaying energy fields within them. Your simple Lithium polymer cell will not cause this - all I would do is to put a 2A fuse in series with the cell positive lead, close to the cell in order to protect from overcurrent scenarios.