How to do last step of Krom's algorithm for solving 2SAT problems
The first step in implementing an algorithm is understanding why it works.

In your case, Krom's algorithm is based on repeated application of the resolution rule
$$
\begin{array}{c}
a \lor b \qquad \bar{a} \lor c \\\hline
b \lor c
\end{array}
$$
Here $a,b,c$ are literals. This rule is valid, and applied to two 2-clauses produces a 2-clause. Hence there is no harm in running it until reaching a point at which it can be applied no more. If in this way we reach a clause $a \lor a$, then the literal $a$ must be true. So if we reach both $a \lor a$ and $\bar{a} \lor \bar{a}$, there is no solution.

More interestingly, if all solutions have $a$ set to true, then the completeness of resolution implies that $a \lor a$ will be generated. 

This suggests the following algorithm:


Generate all 2-clauses reachable by resolution.
If you have generated both $a \lor a$ and $\bar{a} \lor \bar{a}$ for some variable $a$, then there is no solution.
Otherwise, if for some variable $a$ we have generated neither $a \lor a$ nor $\bar{a} \lor \bar{a}$, add one of them as you please and go back to step 1.
Output the truth assignment satisfying all literals $a$ for which we have generated $a \lor a$.


This is just one implementation; others are possible. (For example, we could immediately substitute any known values before step 3.)

Back to your example: $a \lor b$. Step 1 produces no new clauses. The variable $a$ is "free", so in step 3 we add (say) $\bar{a} \lor \bar{a}$. Step 1 is now more interesting:
$$
\begin{array}{c}
a \lor b \qquad \bar{a} \lor \bar{a} \\\hline
b \lor \bar{a}
\end{array}
\\
\begin{array}{c}
a \lor b \qquad b \lor \bar{a} \\\hline
b \lor b
\end{array}
$$
We can now read off the truth assignment $a=F,b=T$.

Had we added $a \lor a$ instead, step 1 would still not do anything, and so we would add either $b \lor b$ or $\bar{b} \lor \bar{b}$ in step 3, and then read off one of the truth assignment $a=T,b=T$, $a=T,b=F$.