What does the Lorentz factor represent? How can the Lorentz factor $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ be understood? What does that mean? For example, what is the reason for the second power and square root? Why not $\frac{1}{1-\frac{v}{c}}$, or what would happen if it took that form?. Can you point me to other physical laws that make use of the $\sqrt{1-r^2}$ so as to "translate" it.
Let $T_0$ be the local period and $L_0$ the local length

*

*Light-clock moves $\bot$ light $\rightarrow$ $T=\frac{T_0}{\sqrt{1-\frac{v^2}{c^2}}}$

*Light-clock moves $\parallel$ light $\rightarrow$ $T=\frac{2L}{c\left(1-\frac{v^2}{c^2}\right)}$
Referring to 1, How could photons not miss the mirror?
Q: How could I understand the Lorentz factor formula intuitively, or what is your concept of it?
 A: The Lorentz factor can be understood as how much the measurements of time, length, and other physical properties change for an object while that object is moving.
What you have named $r^2$ is indeed known as $\beta^2$ which is the ratio between the relative velocity between inertial reference frames and $c$ the speed of light. It is also written as
$$
\gamma = \frac{1}{\sqrt{1-\beta^{2}}}=\frac{d t}{d \tau}
$$
where $t$ is coordinate time and $\tau$ is the proper time for an observer (measuring time intervals in the observer's own frame). This is also what you have written as equation (1) on your question. For the interpretation of (1) don't think of a mirror itself, just think of an infinitely big plane perpendicular to your light beam/direction of the photons.
It is also related to many other quantities as I mentioned when I was defining what the factor is, I will write a few examples below:

*

*Time Dilation: $\Delta t^{\prime}=\gamma \Delta t$

*Length Contraction: $\Delta x^{\prime}=\Delta x / \gamma$

*Relativistic mass: $m=\gamma m_{0}$
As you can see, all this quantities are $\gamma$ dependent and therefore evolve with the square root factor you mentioned!

For more examples, search in Wikipedia, there is probably more of them
A: In my opinion, the time-dilation factor $\gamma$ is best thought of as
the
"hyperbolic cosine" $\gamma=\cosh\theta$,

that is, the "cosine function" (as in the ratio "ADJACENT/HYPOTENUSE" in a Minkowski-right triangle) in Minkowskian hyperbolic-trigonometry [the flat geometry where the "hyperbola" plays the role of the circle].
In a coordinate-free way, the time-dilation factor $\gamma=\hat u \cdot \hat v$
is the Minkowski-dot-product between two 4-velocities.
Why the square-root?
It's similar to why
$$\cos\phi=\frac{1}{\sqrt{1+m^2}},$$
when $\cos\phi$ is expressed in terms of the slope $m=\tan\phi$.
Starting from Pythagoras,
$$\cos\phi=\frac{\Delta x}{\Delta s_{Euc}}=\frac{\Delta x}{\sqrt{\Delta x^2+\Delta y^2}}
=\frac{1}{\sqrt{1+\left(\frac{\Delta y}{\Delta x}\right)^2}}.
$$
So, in Minkowski spacetime,
$$\cosh\theta=\frac{\Delta t}{\Delta s_{Min}}=\frac{\Delta t}{\sqrt{\Delta t^2-\Delta y^2}}
=\frac{1}{\sqrt{1-\left(\frac{\Delta y}{\Delta t}\right)^2}}.
$$
And, in Galilean spacetime, the "galilean cosine" (as defined by I.M. Yaglom)
$${\rm cosg\ }\eta=\frac{\Delta t}{\Delta s_{Gal}}=\frac{\Delta t}{\sqrt{\Delta t^2-0\Delta y^2}}
=\frac{1}{\sqrt{1-0\left(\frac{\Delta y}{\Delta t}\right)^2}}=1,
$$
that is, the "no-time-dilation-factor".
(This sequence of cosine functions [in these affine Cayley-Klein geometries] suggests that our "common sense" Galilean intuition is more the exception, rather than the rule.)
The above can be made more formal and more precise, as some of the other answers show. But I think for the purpose of "developing intuition", I've chosen to make contact with what I think are more familiar facts, treated by analogy.
A: Why does it have a square root? For these kinds of things there is always a bit of historic justification. There could have been a different definition if things had turned out differently. But aside from that the way that $\gamma$ is defined makes it the simplest definition in most cases. For example

*

*Time dilation $\Delta t=\gamma\Delta t_0$

*Length contraction $L=\dfrac{L_0}\gamma$

*Lorentz transformation $x'=\Lambda x$ where $\Lambda=\pmatrix{\gamma&-\gamma\beta\\-\gamma\beta&\gamma}$
What is some intuition behind the form $1/\sqrt{1-\beta^2}$? The only intuition I could come up with is that a unit circle is described by $y=\sqrt{1-x^2}$ which means that you can of $\gamma$ as the inverse of the height of point with $x=\beta$ that lies on a unit circle.
What is my concept of $\gamma$? To me $\gamma$ is used to define the three examples that I mentioned before. Also it measures how strong relativistic effects are. $\gamma=1$ means no relativistic effects and as $\gamma\rightarrow\infty$ relativistic effects become more and more important.
A: Consider the possible ways one could formulate transformation laws for kinematics and coordinates that respect isotropy (that is: the equivalence of the spatial coordinates). A large range of possibilities (but not the largest possible) is captured by the following family:
$$δ = -βt, δt = -α·$$
that describe, in 3-vector form, the transform of the spatial coordinates $ = \left(x,y,z\right)$ and time coordinate $t$ under an infinitesimal boost $$. Transformations that change the motion of the reference system from one inertial motion to another are called "boosts". These are the boosts which leave the following quantity
$$βt^2 - αr^2$$
invariant ... at least in the cases where $\left(α,β\right) ≠ \left(0,0\right)$. In the case $\left(α,β\right) = \left(0,0\right)$, $t$ and $r^2$ are each invariant separately.
Integrate the infinitesimal transforms to finite form as the transforms for a finite boost by setting it up as differential equations
$${d \over dλ} = -βt, {dt \over dλ} = -α·, {d \over dλ} = ,$$
now treating $(λ)$, $t(λ)$ and $(λ)$ as functions of $λ$.
Solve by integrating for $λ = 0$ to $λ = 1$, with the original coordinates being
$$\left(,t\right) = \left((0),t(0)\right)$$
and the transformed coordinates as
$$\left(',t'\right) = \left((1),t(1)\right).$$
The general solution for $$ is just $(λ) = λ$, so we could just take the boost velocity $ = (1)$ to be $$, itself. You can write the transform in terms of the original coordinates and $$.
If $αβ > 0$, then the transform will involve hyperbolic functions (and then $ = $ is called the rapidity, and does not actually coincide with the boost velocity, as you'll soon see); if $αβ < 0$, it will involve circular functions; if $α ≠ 0, β = 0$ you will get the Galilean transforms with $ = $ being the velocity of the boost and $t$ remaining fixed, while for $α = 0, β ≠ 0$, $r^2 = x^2 + y^2 + z^2$ remains fixed. In the last case $α = 0, β = 0$: both $t$ and $r^2$ remain fixed. Those last three cases correspond, respectively, to Absolute Time, Absolute Space and Both. In the case before that, $αβ < 0$, $t$ is just another spatial coordinate and you have a 4-dimensional timeless space, rather than a 3+1 dimensional space-time.
To find what the boost velocity is, start with an inertial motion given by $ = t$ and find the transform that catches up to this motion and makes it stationary: $' = $. This will give you the relation between $$ and $$. If you use that relation to replace $$ by $$, you will recover the Galilei transforms in the cases $α ≠ 0, β = 0$ and the Lorentz transform in the cases $αβ > 0$, as well as finding that $c ≡ \sqrt{β/α}$ is an invariant speed, in that case. Correspondingly, the Galilei case may be regarded as the limit $c → ∞$, while the other unnamed case $α = 0, β ≠ 0$ is the limit $c → 0$. The case $α = 0, β = 0$ is, in effect, both limits at the same time.
You'll notice, by the way, that you generally will not be able to transform down to a rest frame in the cases where $β = 0$. Motion and rest are absolute in those cases and the boost actually gives you something that would be more accurately measured as inverse speed (or marathon-runners speed), which you could probably call "slowness": a set of units being miles per minute, for instance.
You will also find a restriction on $$ in the case $αβ > 0$ that $v < c$. If you start out with any $$ such that $v > c$ then it will stay so as $v' > c$ under transform and it will not be possible to transform to a rest frame. These are not the trajectories of motions at all, but just lines in space. That above restriction in the cases $β = 0$ are limiting cases of this restriction. Finally, any $$ such that $v = c$ will only transform to a $'$ such that $v' = c$. The speed $c$ is absolute.
Adopting the usual ploy of aligning the boost to the $x$ direction, then the solution you should get is:
$$\left(x',y',z',t'\right) = \left({{x - βwt} \over \sqrt{1 - αβw^2}}, y, z, {{t - αwx} \over \sqrt{1 - αβw^2}}\right)$$
with a boost speed $v = βw$ that is only meaningful in the cases $β ≠ 0$. For the cases $β = 0$ and $α ≠ 0$, you'd have to use the measure of "slowness" $∧ = αw$, instead. In the 4-D timeless space case, $αβ < 0$, this does not cover the full range of transforms, since the signs of $x$ and $t$ can be flipped by passing them through a 180 degree circular rotation. But writing it this way helps to show its connection to the other cases.
The rapidity $u$ is related to $w$ as
$$w = {{\tanh {\sqrt{αβ}u}} \over \sqrt{αβ}}$$
if $αβ > 0$,
$$w = {{\tan {\sqrt{-αβ}u}} \over \sqrt{-αβ}}$$
if $αβ < 0$ and
$$w = u$$
if $αβ = 0$.
A: The reason for the squared and square roots is that proper time is the quadratic difference of the time and distance: $\Delta \tau^2 = \Delta t^2 - \Delta x ^2$. This differs from Euclidean distance, which is the quadratic sum of the components. Thus, while the components in Euclidean space can be put in terms of quadratic functions of some parameter ($\Delta x = r\cos(\theta), \Delta y = r\sin(\theta)$), relativistic components can be put in terms of hyperbolic functions: $\Delta x = \Delta \tau \sinh(\eta), \Delta t = \Delta \tau \cosh(\eta)$.  $v$ is the ratio of spatial displacement to the time displacement: $v/c =  \frac{\Delta x}{\Delta t} = \frac{\sinh(\eta}{\cosh \eta}=\tanh(\eta)$.
The "Lorentz factor" is simply the ratio between the proper time and the projection of proper time onto a "rotated" coordinate system. In Euclidean space, a vector projected onto a basis that is at an angle of $\theta$ to it has its length multiplied by $\cos(\theta)$. In Minkowski space, "rotating" by $\eta$ results in a multiplication by $\cosh(\eta)$. And when you write $\cosh(\eta)$ in terms of $\tanh(\eta)$ (that is, $v/c$), you get $\cosh(\eta)=\frac 1 {\sqrt{1-\tanh^2(\eta)}}$, or $\cosh(\eta)=\frac 1 {\sqrt{1-(v/c)^2}}$.
If you put things in terms of $\eta$, many formulae become simpler. For instance, instead of the complicated addition formula you have when combining velocities, combining two $\eta$s requires just adding the $eta$s. And the Doppler effect is just $e^{\eta}$.
A: Just to add to the other answers, one might ask, "why the square root"?
The heart of it is the underlying geometry.  When we ask "what is the diagonal of a square with sides $a$ and $b$, Euclid's rules eventually land us at the conclusion $\sqrt{a^{2} + b^{2}}$, which then comes from the fact that, in a flat Euclidean space, distances are given by $ds^{2} = dx^{2} + dy^{2} + dz^{2}$
For reasons that ultimately come down to "we need a way to differentiate space from time in order to preserve causaility, while also combining the two", it works out that the distance in spacetime is given by (the sign is an arbitrary choice that once made, must be adhered to, but is otherwise arbitrary -- most people in quantum field theory choose the top signs, most in general relativity choose the bottom signs)
$$ds^{2} = \pm(c\, dt)^{2} \mp dx^{2} \mp dy^{2} \mp dz^{2}$$
where, for this purposes, $c$ has to be some "conversion factor" with the units of velocity for this to make sense, and it works out that this has been shown to be the "speed of light".  Now,  choosing the top signs, and noting that $\frac{dt}{dt} =1$, we can divide both sides by $(c\,dt)^{2}$, and the RHS of the above becomes:
$${}^{4}v^{2} = 1 - \left(\frac{{}^{3}v}{c}\right)^{2}$$
where the left-sided superscripts indicate three-and four-dimensional velocities.  We can definitely be a lot more rigorous here, and there are great depths to work through, but if you're to ask "where does the square root and that minus sign come from?", the heart of it is that negative sign in the distance function, which can be taken as a core axiom of special relativity.
A: To spell out @Accumulation's and @AccidentalTaylorExpansion's answers in a slightly different way: $γ$ (the Lorentz factor) is a scaling factor, and applies as part of a hypotenuse.
$$\begin{align}
  γ & = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \\
γ^2 & = \frac{1}{1 - \frac{v^2}{c^2}} \\
    & = \frac{c^2}{c^2 - v^2} \\
\\
γ^2 (c^2 - v^2) & = c^2 \\
c^2 - v^2 & = \frac{c^2}{γ^2} \\
          & = (\frac{c}{γ})^2 \\
c^2 & = v^2 + (\frac{c}{γ})^2
\end{align}$$
The last line shows $c$ is a hypotenuse, and as this ASCII art shows:
c  /|
  / | c/γ
 ---+
   v

the greater $v$ is in comparison to $c$, the smaller will be $c/γ$ (i.e. the larger $γ$) - or the subjective speed of light: this shows the dilation of time (or contraction of space) in special relativity.
A: The answers given so far are great and I can't top those. I do have a couple points that they didn't address, though.
Velocity, $\vec v$, is a vector and $\gamma$ is a scalar. $\frac{\vec v}{c}$ is a vector scaled such that it is unitless, but it still has 3 components (in 3-space). As such, $1 - \frac{\vec v}{c}$ is un-defined, as is $\frac{1}{1-\vec v/c}$. Supposing $\gamma$ had some unitless vector to subtract from, i.e. $\sqrt{\vec a - \frac{\vec v}{c}}$, the square root of a vector is a spinor which doesn't have a unique associated scalar.
$\gamma$ may be written with $\frac{v^2}{c^2}$, but it is implicitly $\frac{\vec v \cdot \vec v}{c^2}$ which is a simple way to convert a vector with units into a unitless scalar (a plain old real number).
When we finish a calculation we always check that we have the correct units. Making sure we also have legal operations and a tensor of the correct rank and size is even more important and just as easy to check.
Examining the limits of a term helps a lot with building intuition
$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
In the limit that $v \ll c$, $\gamma \rightarrow 1$ and we have classical mechanics while $\lim_{v \rightarrow c} \gamma \rightarrow \infty$.
