The importance of the phase in quantum mechanics

When people say that the phase doesn't matter, they mean the overall, "global" phase. In other words, the state $|0 \rangle$ is equivalent to $e^{i \theta} |0 \rangle$, the state $|1\rangle$ is equivalent to $e^{i \theta'} |1 \rangle$, and the state $|0\rangle + |1 \rangle$ is equivalent to $e^{i \theta''} (|0 \rangle + |1 \rangle)$.

Note that "equivalence" is not preserved under addition, since $e^{i \theta} |0 \rangle + e^{i \theta'} |1 \rangle$ is not equivalent to $|0 \rangle + |1 \rangle$, because there can be a relative phase $e^{i (\theta - \theta')}$. If we wanted to describe this very simple fact with unnecessarily big words, we could say something like "the complex projective Hilbert space of rays, the set of equivalence classes of nonzero vectors in the Hilbert space under multiplication by complex phase, cannot be endowed with the structure of a vector space".

Because the equivalence doesn't play nicely with addition, it's best to just ignore the global phase ambiguity whenever you're doing real calculations. Finally, when you're done with the entire calculation, and arrive at a state, you are free to multiply that final result by an overall phase.

The global phase does not matter. In your example $\lambda(\vert\psi_1\rangle+\vert\psi_2\rangle)$ has the same physical contents as $\vert\psi_1\rangle+\vert\psi_2\rangle$ but this will be in general different from $\vert\psi_1\rangle-\vert\psi_2\rangle$ or more generally $\lambda’(\vert\psi_1\rangle+e^{i\varphi}\vert\psi_1\rangle).$

... and of course yes the relative phase can be measured, as indicated for instance in this answer and no doubt many others. In fact interferometry depends on such relative phases.

While the other answers are correct, this is not a different answer but rather an illustration that the relative phase is indeed important in quantum mechanics. We know that bosons (particles with integer spin) have the following property: a rotation by $2\pi$ (around any fixed axis) leaves their states invariant, $R(2\pi)|{\rm boson}\rangle = |{\rm boson}\rangle$. This is obviously fine, since a rotation by $2\pi$ should be a symmetry operation. Fermions (particles with integer-and-a-half spin) have the property that a rotation by $2\pi$ changes their sign: $R(2\pi)|{\rm fermion}\rangle = -|{\rm fermion}\rangle$. This is also fine, since $-|{\rm fermion}\rangle$ belongs to the same ray as $|{\rm fermion}\rangle$ and hence describes the same state.

What, however, if we want to make a linear superposition of the form $|\Psi\rangle = \alpha|{\rm boson}\rangle+\beta|{\rm fermion}\rangle$, with $\alpha\neq\beta$? It is clearly seen that the operation of rotation by $2\pi$ on $|\Psi\rangle$ will not give a state proportional to $|\Psi\rangle$, and so is not a symmetry of that state. What went wrong?

The answer is that we simply should not be making such superposition. While it is well-defined mathematically, it is unphysical: it does not describe a state that can be physically prepared. Thus, we are forbidden from making a (physical) superposition of a boson and a fermion. This is an example of a powerful class of statements known as superselection rules.