What do the gamma matrices mean in QED’s Feynman diagrams? From page 245 of the PDF of Matthew Schwartz's book (page 226 of the textbook), he shows a Feynmann diagram in QED like this:

We can see the spinors/propagator for the electrons and the photons coming in.
I see what the gamma matrices have some kind of meaning here. I think that all they represent are vertices of a Feynman diagram, however, I think there could be more to it than that, and I just want to make sure by asking here.
What I want to know is:

*

*Do they just represent the vertices of a diagram or something more in the context of a Feynman diagram?

*What are the reasons we would use a gamma matrix outside the use of a vertex of a Feynman diagram? A couple of examples would be using them in gradients in Lagrangians (i.e. $\gamma^{\mu}\partial_{\mu}$) and inside propagators with the Feynman slash notation (as shown in the picture above)

 A: The gamma matrix is related to the fact that the electron is spin 1/2, and that they come in both normal and anti-particle types. Let me work through that a little bit.
First, symmetry usually comes in a group. And that means, in this case, representations using matrices. A matrix representation is a set of matrices that have the same multiplication table as the group. If in the group you have three elements $A$,  $B$, and $C$, and if $ A \times B = C$, then you need the matrices that represent this to satisfy the same relation. If $M(A)$ is the matrix repreenting $A$ and so on, then $M(A) \times M(B) = M(C)$.
And you are interested in representations because you want a particle to rotate into itself. Or, at least, a rotated version of itself. You want an electron to still be an electron after you rotate it. A photon to still be a photon. You don't want a mixture of an electron and photon if you started with a photon.
Rotation applied to a vector can be represented by a rotation matrix. In 3-dimensions this is the very familiar thing of cos and sin.
$$ \left( \array{\cos \theta & - \sin\theta & 0 \\  \sin \theta  &  \cos\theta & 0 \\
0 & 0 & 1} \right) $$
This, of course, is a rotation around the z-axis in a 3-D system. You can easily make rotations around the x-axis or y-axis by using cos/sin in the appropriate locations. And then matrix products can build up arbitrary rotations. Putting it all together gives you SO(3), special orthoganal 3x3 matrices.
When the thing getting rotated is a spin 1/2 object, some more complication enters. For a spin 1/2 particle that you don't worry about particle-anti-particle, you get the Pauli Spin matrices. Here is one representation of them.
$$ \sigma_1 = \left( \array{ 0 & 1 \\ 1 & 0} \right) $$
$$ \sigma_2 = \left( \array{ 0 & -i \\ i & 0} \right)  $$
$$ \sigma_3 = \left( \array{ 1 & 0 \\ 0 & -1} \right)  $$
And with some rather involved maths that I won't reproduce here but that are in the link, you can show that these can be the generators of a 2x2 representation of the the group SO(3).
You can motivate this a little, without actual proof, as follows. Consider the circle defined by one complete rotation of $\theta$ in the 3x3 rotation matrix. Now think of that as a circle of matrices, each with a particular value of $\theta$. You get a matrix-valued circle. Now think about the tangent to that circle at each value of $\theta$. The Lie group is the circle. (Sort of.) The Lie algebra is the tangent.  (Again, sort of.)
So you go off and spend a year studying group theory and representations in terms of matrices and Lie groups and Lie algebras. And one thing you find is that there is a matrix representation for each size of matrix.
The 1x1 matrix is just scalars. Rotations of scalars are kind of boring, they just rotate onto themselves.
The 3x3 rep is the spin-1 rep. And going round 360 degrees brings you back. Yay! It behaves like you or me on an office chair.  When there are anti-particles it becomes a 4x4 rep, and it becomes photons. That's the $A_\mu$ in your equation.
But the Pauli spin matrix rep requres 720 degrees to get back to its starting point.  Here you need to know about the Dirac belt trick. There are easy to construct physical systems that can return to their origin only after 720 degrees.
And electrons are like that. You need two full rotations to get back. And the Pauli spin matrices are intended to construct just such a matrix representaion. Basically, the wave function for such a spinor is an ordered collection of two complex numbers. That's just a pedantic way of saying it's got a spin-up part and a spin-down part, and we write them as a pair. So we need a 2x2 matrix to act on it to rotate it.
But! But electrons in relativisitc quantum mechanics have anti-particles. That means an electron has four components: spin-up and down for each of particle and anti-particle. And that complicates the situation a bit. Now we need 4x4 matrices to represent the rotations of particle and anti-particle parts of the spinor. In the equations you wrote, the $\gamma^\mu$ are the following.
$$ \gamma^0 =  \left( \array{
1 & 0 & 0 & 0 \\
0 &1 & 0 & 0 \\ 
0 & 0 & -1 &0 \\
0 & 0 & 0 &-1} \right) $$
$$ \gamma^1 =  \left( \array{
0 & 0 &0 & 1 \\
0 & 0 & 1 & 0  \\ 
0  & -1 & 0 & 0  \\
-1 & 0 & 0 & 0} \right) $$
$$ \gamma^2 =  \left( \array{
0 & 0 &0 & -i \\
0 & 0 & i & 0  \\ 
0  & i & 0 & 0  \\
-i & 0 & 0 & 0} \right) $$
$$ \gamma^3 =  \left( \array{
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1  \\ 
-1  & 0 & 0 & 0  \\
0 & 1 & 0 & 0} \right) $$
You have three matrices that correspond to the Paul spin matrices, but in a 4x4 rep of the group. And you get an extra matrix because you now have to include particle-anti-particle in the system.
And to wrap up the whole thing, the reason the $\gamma_\mu$ shows up where it does is because it connects the vector rep of the photon, the $A_\mu$ to the spinor rep of the electron. When an electron interacts with a photon, part of what it does is change the phase of the electron. And, in a certain sense, this can be thought of as a rotation. It basically means that when an electron, a spin 1/2 particle, emits a photon, a spin 1 particle, it has to rotate its spin from up to down. That's the only way it can conserve angular momentum.
A: To see where gamma matrices come from, you should find a textbook which covers relativistic quantum mechanics and Dirac Equation. One of the textbooks is Sakurai's Modern Quantum Mechanics.
Since you are reading a quantum field theory book, there has to be a part about quantization of fermionic fields. In order to quantize fermions, we should first find out the classical solutions of Dirac Equation, which involve gamma matrices and are called spinors.
Essentially, these gamma matrices are introduced first when Dirac first tried to include special relativity into quantum mechanics. It turned out to be a complicated business and one cannot describe fermions as scalars, otherwise it is impossible to write down a quantum equation that's consistent with the relativistic dispersion relation $E^2=p^2+m^2$. It turned our that we have to use as least 4-component spinors, and the equation governing it is Dirac Equation, which has these gamma matrices acting on spinors.
As for why gamma matrices appear in this particular Feynman diagram, one should remember that the amplitude of such a diagram represent the effect of some interaction on the scattering amplitudes, in this case, the interaction Lagrangian is written as
$$
\mathcal{L}_{int}=-e\bar{\psi}\gamma^{\mu}A_{\mu}\psi
$$
which has gamma matrices in it. In Feynman diagrams, interactions are represented by internal vertices. This particular interaction involve two electrons and one photon
One could, of course, ask why the QED interaction should involve such an interaction term with gamma matrices in them, the answer is, the most natural way of coupling electrons with photons is by changing the Free fermionic theory + Free Photon theory
$$
\mathcal{L}_{free}=\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}
$$
Into the following interacting theory
$$
\mathcal{L}=\bar{\psi}\left(i\gamma^{\mu}D_{\mu}-m\right)\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}
$$
where
$$
D_{\mu}\equiv\partial_{\mu}+ieA_{\mu}
$$
is the gauge covariant derivative.
You can plug this in $\mathcal{L}$ to check that this does produce an interaction term $\mathcal{L}_{int}=-e\bar{\psi}\gamma^{\mu}A_{\mu}\psi$.
This Gauge covariant derivative also appears when we try to couple ordinary quantum mechanics with EM field. For magnetic quantum systems we change our momentum $\mathbf{p}$ into $\mathbf{p}+e\mathbf{A}$, where we have a vector field and has a gauge choice. This new derivative is defined such that it transforms nicely under gauge transformation and makes quantum mechanics under different gauge equivalent up to a local phase change.
A: Feynman diagrams are shorthand for some integrals in quantum field theory. Gamma-matrices are also used in the Dirac equation.
EDIT (Dec. 27, 2021): Now the OP asks: what are the reasons we would use a gamma matrix outside the use of a vertex of a Feynman diagram. It turns out that gamma-matrices represent basis vectors in the Minkowski spacetime and also reflections there (a gamma-matrix acting on a spinor produces a reflected spinor - see Lecons sur la theorie des spineurs, 1938, by E. Cartan, the discoverer of spinors.  I believe there is an English translation).
