Trying to understand this measurement of a simple quantum circuit

You can think of the circuit operation as follows. Remembering that for Hadamard matrix: $H times H = mathbb{I}$, your circuit looks like the following: $$ H otimes mathbb{I} otimes mathbb{I}(CCNOT(H|0rangle otimes H|0rangle otimes |0rangle )) $$ Let's focus on the inner-most part first: $$ H|0rangle otimes H|0rangle otimes |0rangle =frac{1}{sqrt{2}}(|0rangle + |1rangle) otimes frac{1}{sqrt{2}}(|0rangle + |1rangle) otimes |0rangle = frac{1}{2}(|000rangle + |010rangle + |100rangle + |110rangle). $$ Now, the action of $CCNOT$ gate is that, if the first two qubits are $1$, then the third qubit would be flipped. But notice that in the above state, only the last term has two $1$'s in it, i.e. the $|110rangle$ term. So, the action of $CCNOT$ on the above state would be: $$ CCNOT(frac{1}{2}(|000rangle + |010rangle + |100rangle + |110rangle)) = frac{1}{2}(|000rangle + |010rangle + |100rangle + |111rangle). $$ Now we apply the Hadamard matrix on the first qubit only: $$ H otimes mathbb{I} otimes mathbb{I}(frac{1}{2}(|000rangle + |010rangle + |100rangle + |111rangle)) = frac{1}{2}(frac{1}{sqrt{2}}(|0rangle + |1rangle)|00rangle + frac{1}{sqrt{2}}(|0rangle + |1rangle)|10rangle + frac{1}{sqrt{2}}(|0rangle - |1rangle)|00rangle + frac{1}{sqrt{2}}(|0rangle - |1rangle)|11rangle)). $$ Simplify it! Would be a good exercise if nothing else. :) You would see that the resulting state has the terms that you have shown in the second picture.