We can solve this using the method of recursion. We have three cases to consider. First, we throw two $H$s on our first two throws, giving us a result of 2. Second, our first throw is a $T$. In this case, we must add one to our expectation. Third, we throw $HT$. In this case, we must add two to our expectation. So, our equation is $$ \text{E}[HH] = 2 P(HH) + P(T) \left( \text{E}[HH] + 1 \right) + P(HT) \left( \text{E}[HH] + 2 \right). $$ It is straightforward to see that $$ P(T) = \frac{1}{2} $$ $$ P(HH) = \frac{1}{4} $$ $$ P(HT) = \frac{1}{4} $$ and solving the above equation gives $$ \frac{1}{4} \text{E}[HH] = 1.5 $$ $$ \text{E}[HH] = 6. $$