Quantum coherence isn't a measure that applies to the states $|\phi_1⟩$ and $|\phi_2⟩$ themselves, but rather to superpositions of the two. Basically, quantum coherence is what separates a true superposition state like
$$
|\psi_+⟩ = \frac{1}{\sqrt{2}}(|\phi_1⟩+|\phi_2⟩)
$$
from a classical mixture where you just flip a fair coin and prepare $|\phi_1⟩$ or $|\phi_2⟩$ depending on the coin-toss outcome.
The basic question is this: suppose that I'm a salesman, and I'm trying to sell you a box which (I claim) contains a Schrödinger cat in the superposition state |dead⟩+|alive⟩. One way that this can be verified is by opening the box (or rather, an ensemble of such boxes) and confirming that once they've been opened, the boxes contain 50% dead cats and 50% live cats: this is useful data, but it is not enough to tell whether I'm actually selling you superpositions or whether I'm a scammer and all I did was to flip coins and just put half the cats in a pre-dead state.
To be able to tell the difference, what you really need is a second, incompatible measurement. Say, for example, that you have a system that is presumably in the $|\psi_+⟩$ state from above, and you've been able to verify that a measurement on the $\{|\phi_1⟩,|\phi_2⟩\}$ basis produces a 50:50 split of both outcomes, and you want to go beyond that and conclusively distinguish $|\psi_+⟩$ from a classical probabilistic mixture. One definite silver bullet for this is if you can perform a quantum projective measurement in the basis $\{|\psi_+⟩,|\psi_-⟩\}$, where
$$
|\psi_-⟩ = \frac{1}{\sqrt{2}}(|\phi_1⟩-|\phi_2⟩),
$$
because it takes only a bit of algebra to show that a system prepared in $|\psi_+⟩$ and measured in that basis will never produce the outcome $|\psi_-⟩$. This is part of a wider pattern, in that if you can measure in the basis
$$
\left\{
|\psi_+(\varphi)⟩ = \frac{|\phi_1⟩+e^{i\varphi}|\phi_2⟩}{\sqrt{2}}, \
|\psi_-(\varphi)⟩ = \frac{|\phi_1⟩-e^{-i\varphi}|\phi_2⟩}{\sqrt{2}}
\right\},
$$
having prepared the system in the state $|\psi_+⟩$, then the probability of obtaining $|\psi_+(\varphi)⟩$ will be a sinusoidal interference pattern,
$$
P_+(\varphi)
=
\left|⟨\psi_+(\varphi)|\psi_+⟩\right|^2
=
\frac12 + \frac12 \cos(\varphi).
$$
The quantum coherence of the system, in this situation, refers to its ability to produce such an interference pattern, both in the qualitative sense of whether fringes show up or not, as well as in the quantitative sense of how visible they are: as a concrete example, a system with finite coherence $q$ (where $0\leq q\leq 1$) would produce an interference fringe pattern of the form
$$
P_+(\varphi)
=
\frac12 + \frac12 q \cos(\varphi)
$$
under that same measurement, where the interference fringes are damped by a factor of $q$.
The scammer example from above is the $q=0$ extreme, where there is simply no dependence on $\varphi$, and the system is not in anything that you'd remotely begin to call a superposition state. However, it is also possible for a system to have imperfect but nonzero coherence, with $0<q<1$, with the fringes still visible but not spanning the totality of their allowed range.
And obviously, if you want to do nice QM experiments with interference effects, you want things to be as coherent as possible.
Here I should mention that on a more technical sense, the coherence of a quantum system is always relative to a specific basis (i.e. different measurements might make the system look somewhat more or less coherent, though a related quantity called the purity is basis-independent) and that on more technical language it is given by the off-diagonal elements of its density matrix. However, at your stage those technical precisions are probably secondary to the conceptual exposition above.
By true superposition state do you actually mean $\alpha = \beta = \frac{1}{\sqrt{2}}$ in $\alpha|\phi_1>+\beta|\phi_2>$. Other $\alpha$'s and $\beta$'s can be superposition too.
– Jitendra Aug 24 '18 at 15:48