As @Prahar noted in the comments, there's no problem defining a four velocity for light. The resulting four vector however will be a light-like four vector, so it won't have the same normalization as a four velocity of a massive particle. For massive particles $u^\mu u_\mu = -c^2$ (East coast metric signature assumed), but for photons which are massless we'll get $k^\mu k_\mu = 0$. From now on however, let's use units where $c=1$ for simplicity.
Your question touches on several interesting issues, and since I'm not sure to what extent you want to get into each, I'll start with the most obvious things that I'm quite sure you're asking about, and end with things that may be of less interest but that I think are still related. If you find them unhelpful now, perhaps they can help you in the future as you get deeper into the subject.
Four velocity for light - working out the definition
From your question, I'm not sure if you are familiar with the subject of four momentum and relativistic dynamics at all yet. Nevertheless, for the purposes of this answer, take it as a given that we can write a four momentum vector for a photon as follows:
$$ \mathbf{p}_\gamma = \left(E_\gamma,\vec{p}_\gamma \right), $$
It turns out that the four momentum vector of any single particle has the nice property that squaring it (dotting it with itself) results in the mass squared of the corresponding particle (up to a sign and factors of $c^2$). So since for a photon $m_\gamma=0$ we have:
$$ \mathbf{p}_\gamma^2 = -E_\gamma^2+\vec{p}_\gamma^{\ 2} = 0,$$
which also gives us the expected result for a photon which I hope you've seen somewhere before, $E_\gamma=|\vec{p}_\gamma|$ (often written as just $E=pc$). We also see that it enables us to rewrite the four momentum as:
$$ \mathbf{p}_\gamma = \left(E_\gamma,E_\gamma\hat{n} \right), $$
where $\hat{n}=\large\frac{\vec{p}_\gamma}{|\vec{p}_\gamma|}$ is a unit vector pointing along the direction of motion of the photon.
Side note: don't confuse subscript $_\gamma$ with the Lorentz factor. The subscript $_\gamma$ is a standard way to denote quantities belonging to a photon.
Now indeed, to define the four velocity for the photon, a natural way would be to divide the above expression for $\mathbf{p}_\gamma$ by an appropriate factor, that has the dimensions of mass. The natural choice is $E_\gamma$, since $E_\gamma$ has units of mass (remember, when $c=1$ all velocities are dimensionless, so energy and mass have the same dimensions). Therefore we define:
$$ \mathbf{u}_\gamma = \frac{1}{E_\gamma}\left(E_\gamma,E_\gamma\hat{n} \right) = \left(1, \ \hat{n}\right).$$
Now clearly, $\mathbf{u}_\gamma^2=0$ and we've lost the time-like normalization, which is to be expected. The worldline of a photon is light-like.
The composition of velocities (and of happy coincidences)
If I am reading the first part of your question correctly, you already figured out that if you have two observers moving with four velocity vectors written relative to some frame:
$$ \mathbf{u}_1 = \gamma_1(1,\vec{v}_1) \quad \mathbf{u}_2 = \gamma_2(1,\vec{v}_2), $$
then to find the relative gamma factor $\gamma_{\text{rel}}$ between both observers, you can apply the Lorentz invariance of the dot product and obtain:
$$ -\gamma_{\text{rel}} = -\gamma_1\gamma_2 +\gamma_1\gamma_2\vec{v}_1\cdot \vec{v}_2, $$
then you can solve for $v_\text{rel}$ associated with $\gamma_{\text{rel}}$ and thus recover the velocity composition formula.
This is a straightforward algebraic exercise, but I'm going to pursue it here at least to a limited degree, because there's a conceptual subtlety involved which is related to your question. Note that the above expression can be written by explicitly expressing the gammas:
$$ \frac{1}{\sqrt{1-v_{\text{rel}}^2}} = \frac{1-\vec{v}_1\cdot\vec{v}_2}{\sqrt{1-v_1^2}\sqrt{1-v_2^2}} \tag{1} $$
At this stage, it is clear that putting either $v_1=1$ or $v_2=1$ makes the equation undefined. However, notice what happens when we go on and solve for $v_\text{rel}$:
$$ \sqrt{1-v_{\text{rel}}^2} = \frac{\sqrt{1-v_1^2}\sqrt{1-v_2^2}}{1-\vec{v}_1\cdot\vec{v}_2} \tag{2} $$
$$ 1-v_{\text{rel}}^2 = \frac{(1-v_1^2)(1-v_2^2)}{(1-\vec{v}_1\cdot\vec{v}_2)^2} \tag{3} $$
$$v_{\text{rel}}^2 = \frac{(1-\vec{v}_1\cdot\vec{v}_2)^2-(1-v_1^2)(1-v_2^2)}{(1-\vec{v}_1\cdot\vec{v}_2)^2} \tag{4} $$
Let us at this point make one simplifying assumption that: $\vec{v}_1\cdot\vec{v}_2 = v_1v_2$. Meaning, both velocities point along the same axis. You can obtain a more general formula for three dimensions, but this will only distract us from the point here:
$$v_{\text{rel}}^2 = \frac{\require{cancel}\cancel{1+v_1^2v_2^2}-2v_1v_2\cancel{-1}+v_1^2+v_2^2\cancel{-v_1^2v_2^2}}{(1-v_1v_2)^2} \tag{5} $$
$$\boxed{v_{\text{rel}} = \frac{v_1-v_2}{1-v_1v_2}} \tag{6} $$
Now here's what I wanted to point out: the fact that we can plug in $(6)$ either $v_1=1$ (or $v_2=1$) and get:
$$v_{\text{rel}} = \frac{1-v}{1-v} = 1, $$
seems to me to be nothing more than a happy coincidence. Note that the formula is still not defined for $v_1=v_2=1$, but strictly speaking, we shouldn't be allowed to put $v_1=1$ nor $v_2=1$ because the $\gamma$ factors are undefined for these velocity values. In fact, if we take for example $v_1=1$, the corresponding four velocity $\mathbf{u}_1$ becomes undefined, which is a basic starting point of the entire derivation.
What was the point where this division by zero issue was sweeped under the rug? It's the transition from $(1)$ to $(2)$ of course, when we inverted the expression for the $\gamma$'s, thus allowing one of them to be $0$, since the denominator after the inversion is now $0$ only when $v_1v_2=1$, and that happens only when $v_1=v_2=1$ (Note that the speeds here are always nonnegative, as they are the magnitudes of the respective vectors $\vec{v}_1, \ \vec{v}_2$).
Proper time, worldline for light, and the relation to its four velocity
Now about the issue of proper time, this becomes somewhat of a non-issue because as shown, you can define a four velocity for light. The thing that is true however, is that you can't parameterize the worldline of a photon by proper time, which is indeed related to the fact that you can't define a $\gamma$ for it. But the parameterization of worldlines is a separate topic that I don't think you meant to ask about.
I'll just mention though, that if you do have a parameterized worldline for light, it is going to be parameterized by some affine parameter usually denoted $\lambda$ which is not proper time of course. Then another way to define the four velocity for light will be $u^\mu = \large\frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda}$. It is interesting to note the freedom you have in choosing the affine parameter for the worldline: if you pick for example $\sigma=2\lambda$ it will be just as good a parameter, but that in turn implies that the four velocity for light is unique only up to a scaling factor. This is going to be of importance in what comes next.
Limited usefulness of the four velocity of light
Now, we're going to attempt something that isn't going to work, and considering what we've seen in the previous sections, we shouldn't be surprised that it won't work. So, just to try and learn something from this, let's see what happens if we take two four velocity vectors and try to compose their corresponding three velocities, by the same trick of using the Lorentz invariance of the dot product we applied before. We'll denote as $\mathbf{u}_\gamma$ the four velocity of a photon and as $\mathbf{u}$ a four velocity of some massive particle:
$$ \mathbf{u} = \gamma(1,\vec{v}) \quad \mathbf{u}_\gamma = (1,\hat{n}), $$
now here comes the important part: normally to figure out the $\gamma_{\text{rel}}$ between the particles, we pick the coordinates where one of them is at rest so that its velocity can be written as $(1,\vec{0})$. However, photons have no rest frame. Note also that $(1,\vec{0})$ is not a light-like vector, hence can't be a photon's four velocity.
So, in this case we must pick coordinates where the massive observer/particle that moves with four velocity $\mathbf{u}$ is at rest. In this frame then, denoted by primes, we have:
$$ \mathbf{u}' = (1,\vec{0}) \quad \mathbf{u}_\gamma' = (1,\hat{n}'), $$
now, using a red color for the equations to emphasize their wrongness, we attempt to use the invariance of the dot product:
$$ \color{#CB4154}{\mathbf{u}\cdot\mathbf{u}_\gamma=\mathbf{u}'\cdot\mathbf{u}_\gamma'} $$
$$ \color{#CB4154}{-\gamma +\gamma \vec{v}\cdot\hat{n} = -1 \tag{7}} $$
$$\color{#CB4154}{ \frac{1- \vec{v}\cdot\hat{n}}{\sqrt{1-v^2}} = 1 }$$
now clearly this can't be right. For suppose it happens that $\vec{v}$ is aligned with the direction of the photon so $\vec{v}\cdot\hat{n}=v$:
$$\color{#CB4154}{ \frac{1- v}{\sqrt{1-v^2}} = 1} $$
$$\color{#CB4154}{ \frac{\sqrt{1- v}}{\sqrt{1+v}} = 1,} $$
this can't hold for all $v$, and is clearly wrong.
What we mainly learn from this is that it doesn't work, repeating again just for emphasis. But, this last expression we have obtained is interesting in some way. It may remind you of a factor you encountered if and when you learned about the relativistic Doppler effect, and if so that's not a coincidence.
A final (but hopefully useful) digression - why is the four momentum much more useful than the four velocity?
If we perform a similar calculation, but now taking the four momentum of the photon instead of its four velocity, we can write:
In a "lab" frame where the observer(/massive particle) is moving relative to us and of course, also relative to the photon:
$$ \mathbf{u} = \gamma(1,\vec{v}),\quad \mathbf{p}_\gamma=(E_\gamma,E_\gamma\hat{n}) $$
In the frame where the observer moving with $\mathbf{u}$ is at rest:
$$ \mathbf{u}' = (1,\vec{0}), \quad \mathbf{p}_\gamma'=(E'_\gamma,E'_\gamma\hat{n}') $$
$$ \mathbf{p}_\gamma\cdot\mathbf{u} = \mathbf{p}_\gamma'\cdot\mathbf{u}' $$
$$ \Rightarrow -\gamma E_\gamma + \gamma E_\gamma \vec{v}\cdot\hat{n} = -E_\gamma'\tag{8} $$
$$ \frac{E_\gamma'}{E_\gamma} = \frac{1-\vec{v}\cdot\hat{n}}{\sqrt{1-v^2}}, $$
if we further note Planck's relation between a photon's energy and its frequency $E_\gamma=h\nu$ then this becomes the familiar relativistic doppler effect formula, relating the frequency of a photon the observer measures $\nu'_{\text{obs}}$ to the frequency $\nu_{\text{lab}}$ as measured in some relatively moving frame we call here "lab":
$$ \frac{\nu'_{\text{obs}}}{\nu_{\text{lab}}} = \frac{1-\vec{v}\cdot\hat{n}}{\sqrt{1-v^2}}. $$
Now the important point here pertaining to your question, is to ask why is it that we got such a useful result from computing the inner product between the four momentum of a photon and a time-like four velocity, but when we dotted the four velocity of a photon with a time-like four velocity we got nonsense?
This has to do with what we've mentioned two sections before this one: the four velocity of light is defined only up to scale. As such we may well say that it has no meaningful scale. However, it is related to the four momentum by the explicit scaling factor which is the energy $E_\gamma$. This is what gives equation $(8)$ the dimensions of energy, that is lacking in equation $(7)$, hence resulting in a non-sensical result.
As I think I cautioned at the outset, I may have gotten too far afield at this point. Nonetheless, I hope you'll find this answer useful at least to some degree!
