Discrepancy between heads and tails
Your coin flips form a one-dimensional random walk $X_0,X_1,\ldots$ starting at $X_0 = 0$, with $X_{i+1} = X_i \pm 1$, each of the options with probability $1/2$. Now $H_i = |X_i|$ and so $H_i^2 = X_i^2$. It is easy to calculate $E[X_i^2] = i$ (this is just the variance), and so $E[H_i] \leq \sqrt{E[H_i^2]} = \sqrt{i}$ from convexity. We also know that $X_i$ is distributed roughly normal with zero mean and variance $i$, and so you can calculate $E[H_i] \approx \sqrt{(2/\pi)i}$.

As for $E[\max_{i \leq n} H_i]$, we have the law of the iterated logarithm, which (perhaps) leads us to expect something slightly larger than $\sqrt{n}$. If you are good with an upper bound of $\tilde{O}(\sqrt{n})$, you can use a large deviation bound for each $X_i$ and then the union bound, though that ignores the fact that the $X_i$ are related.

Edit: As it happens, $\Pr[\max_{i \leq n} X_i = k] = \Pr[X_n = k] + \Pr[X_n = k+1]$ due to the reflection principle, see this question. So
$$
\begin{align*}
E[\max_{i \leq n} X_i] &amp;= \sum_{k \geq 0} k(\Pr[X_n = k] + \Pr[X_n = k+1]) \\
&amp;= \sum_{k \geq 1} (2k-1) \Pr[X_n = k] \\ &amp;=
\sum_{k \geq 1} 2k \Pr[X_n = k] - \frac{1}{2} + \frac{1}{2}\Pr[X_n = 0] \\ &amp;= E[H_n] + \Theta(1),
\end{align*}
$$
since $\Pr[H_n = k] = \Pr[X_n = k] + \Pr[X_n = -k] = 2\Pr[X_n = k]$. Now
$$ \frac{\max_{i \leq n} X_i + \max_{i \leq n} (-X_i)}{2} \leq \max_{i \leq n} H_i \leq \max_{i \leq n} X_i + \max_{i \leq n} (-X_i), $$
and therefore $E[\max_{i \leq n} H_i] \leq 2E[H_n] + \Theta(1) = O(\sqrt{n})$. The other direction is similar.