explained comment

The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$

And since $$|+\rangle$$ is $$\frac{1}{\sqrt 2}\left(|0\rangle + |1\rangle \right)$$,

you can work out that $$H(|+\rangle) = |0\rangle$$

So, $$CNOT(H|+\rangle \otimes |+\rangle)$$ $$= CNOT(|0\rangle \otimes |+\rangle)$$ $$= |0\rangle \otimes |+\rangle$$

You can also check that $$H^2 = I$$ or that the Hadamard gate is both Unitary and Hermitian. $$H = H^\dagger$$ $$H^\dagger = H^{-1}$$ So, $$H = H^{-1}$$, the Hadamard gate is its own inverse.

What you have done is not projection of $$|00\rangle$$ to get the state $$|\phi^+\rangle$$, but you just applied the unitary that takes the computational basis to the Bell basis.

As you said in the comments, true, if you measure a state in a basis, you will get one of the basis vectors as outcomes with different probabilities. To see that, express the state in hand in the measurement basis.

For ex:

$$|00\rangle = \frac{1}{\sqrt 2} (|\phi^+\rangle + |\phi^-\rangle)$$ so you will get $$|\phi^+\rangle$$ with 50% probability and $$|\phi^-\rangle$$ with 50% probability.

Similarly, on expressing $$|++\rangle$$ in the Bell basis as: $$\frac{1}{\sqrt 2}(|\phi^+\rangle + |\psi^+\rangle)$$ you get each of those states with 50% probability on measuring.

The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$

And since $$|+\rangle$$ is $$\frac{1}{\sqrt 2}\left(|0\rangle + |1\rangle \right)$$,

you can work out that $$H(|+\rangle) = |0\rangle$$

So, $$CNOT(H|+\rangle \otimes |+\rangle)$$ $$= CNOT(|0\rangle \otimes |+\rangle)$$ $$= |0\rangle \otimes |+\rangle$$

You can also check that $$H^2 = I$$ or that the Hadamard gate is both Unitary and Hermitian. $$H = H^\dagger$$ $$H^\dagger = H^{-1}$$ So, $$H = H^{-1}$$, the Hadamard gate is its own inverse.