How to connect large traces to pads in PCB?
There are two values you need to worry about: voltage drop, and power dissipation. Both are simple Ohm's Law and are functions of the trace resistance.

The trace resistance is a product of its cross sectional area, and its length.

Reduce the length and you reduce the resistance. Reduce the width and you increase the resistance.

So you can have a shorter run of a narrower trace and still handle the current.

The formula for calculating the resistance of a trace is:

$$
R = \rho \frac{l}{A} \cdot (1 + (\alpha \cdot \Delta T))
$$


\$\rho\$ is the resistivity, which for copper is \$1.68×10^{-8} \Omega/m\$.
A is the cross-sectional area in m²
l is the trace length in m
\$\alpha\$ is the temperature coefficient, which for copper is 0.003862 at 20°C.
\$\Delta T\$ is the temperature difference from 20°C


So for a 300 thou (7.62mm) trace at 1oz, which is a thickness of 0.0347mm, a rectangular cross-section would be 

$$
0.00762 \times 0.0000347 = 0.000000264m²
$$

Of course, with etching and other factors it won't be as thick, nor perfectly rectangular, so reduce that a little - let's say for the sake of convenience it's 0.0000002m².

Then you have a trace that's 0.05m long (5cm).  What is the resistance of that trace at, say 23°C?

$$
R = 1.68×10^{-8}\frac{0.05}{0.0000002} \cdot (1 + (0.003862 \times 3))
$$
$$
R = 1.68×10^{-8} \times 250000 \times 1.011586
$$
$$
R = 0.00425\Omega
$$

So once you have the resistance, and you know the current, you can apply simple Ohm's Law to it.  Say 15A, your upper value.

The voltage dropped across that trace is 
$$
V=IR = 15 \times 0.00425 = 0.064V
$$

The power dissipation will be 
$$
P=I^2R = 15 \times 15 \times 0.00425 = 0.956W
$$

So now you can calculate what the voltage drop and power dissipation would be over your small traces to see if it's tolerable.

There are also various tricks you can employ for handling larger currents.  One of the most common (and old-school) is to leave the traces unmasked, then flood them with extra solder.  This massively increases the cross-sectional area thus reducing the resistance.  You can also use electro-plating to achieve a similar result, though this is considerably harder to do, especially in just a small area of the board.

Using wires instead of (or as well as) traces can also be done.

As an aside, you should also consider if the connections, and the pins used in your connectors, are suitable for carrying up to 15A.