Maximum current draw from a step-up regulated output

The input must not affect the output, it's a closed loop system, which means that in wide range of input voltage, it will regulate the output to what you configured.

And since i haven't seen such unregulated DC/DC for a long time, i suggest to first understand how do you connect it and what are you measuring. Specifically, you have to measure on the module itself to ensure regulation. The wires must be thick enough to hold charge current. Of course, in the end the measurement should be taken near the phone. If the cable is poor, you may need to raise the DC/DC 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.