How is H(jw) = jw correspond to a slope on bode plot?
For a transfer function \$H(j\omega) = j\omega\$, we can clearly see that \$|H| = \omega\$, and \$\arg{H} = \frac{\pi}{2}\$. For a magnitude plot, we care about \$|H|\$.

Remember that a Bode plot has axes showing logarithmic frequency and logarithmic gain; moreover note that a decade (factor of 10) change in a value corresponds to a 20-dB change on the decibel scale. 

Because the magnitude of \$H\$ is simply \$\omega\$, multiplying frequency by 10 (a decade) will multiply the magnitude response by a factor of 10 as well (which corresponds to adding 20 dB). Thus it's evident that the slope of this Bode plot will be 20 dB/decade.

The reason you take the intersection at \$\omega = 1\$ is simply because \$|H(j1)| = 1\$, and unity gain is the same thing as zero decibels of gain.