Two virtual photons emitted and reabsorbed by the electron, one enclosed inside the other. Four vertices on the main fermion line with pairing $(v_1,v_4)(v_2,v_3)$: the outer photon spans the full diagram while the inner photon is contained within it.
The computational chain
From diagram to number—every step of the amplitude calculation.
Nested topology
$$v_1 \xrightarrow{\;p-k\;} v_2 \xrightarrow{\;p-k-l\;} v_3 \xrightarrow{\;p-k\;} v_4$$
The fermion line passes through four vertices in order $v_1, v_2, v_3, v_4$. The outer photon connects $v_1$ to $v_4$ and carries loop momentum $k$; the inner photon connects $v_2$ to $v_3$ and carries loop momentum $l$. The inner photon arc is entirely enclosed within the outer one — this is the defining feature of the nested (as opposed to overlapping) topology. The diagram is 1PI: cutting any single internal line does not disconnect it.
Feynman rules expression
$$-i\Sigma_{\text{nest}}(p) = (-ie)^4 \int\frac{d^dk}{(2\pi)^d}\frac{d^dl}{(2\pi)^d}\;\gamma^\mu\,\frac{\not{p}-\not{k}+m}{(p-k)^2-m^2}\,\gamma^\nu\,\frac{\not{p}-\not{k}-\not{l}+m}{(p-k-l)^2-m^2}\,\gamma_\nu\,\frac{\not{p}-\not{k}+m}{(p-k)^2-m^2}\,\gamma_\mu\;\frac{1}{k^2\,l^2}$$
Reading along the fermion line: vertex $v_1$ contributes $\gamma^\mu$, then the fermion propagator with momentum $p-k$, then vertex $v_2$ contributes $\gamma^\nu$, then the fermion propagator with momentum $p-k-l$, then vertex $v_3$ contributes $\gamma_\nu$, then the fermion propagator with momentum $p-k$ again, then vertex $v_4$ contributes $\gamma_\mu$. The two photon propagators contribute $1/k^2$ (outer) and $1/l^2$ (inner). There are five propagator denominators in total: two fermion propagators with momentum $p-k$, one fermion propagator with momentum $p-k-l$, and two photon propagators $1/k^2$ (outer) and $1/l^2$ (inner). Of these, two depend on $l$: the middle fermion propagator $(p-k-l)^2-m^2$ and the inner photon $l^2$.
The subdivergence: the inner loop
$$\Sigma_{\text{inner}}(p-k) = -e^2\int\frac{d^dl}{(2\pi)^d}\;\gamma^\nu\,\frac{\not{p}-\not{k}-\not{l}+m}{(p-k-l)^2-m^2}\,\gamma_\nu\;\frac{1}{l^2}$$
Hold the outer momentum $k$ fixed and examine the $l$-integration over vertices $v_2$ and $v_3$ alone. This is exactly the one-loop electron self-energy $\Sigma^{(1)}$ evaluated at external momentum $p-k$. By power counting it diverges as $\sim 1/\epsilon$ in $d = 4-2\epsilon$ dimensions. This is a subdivergence: the full two-loop integral is ill-defined even before we consider the $k$-integration, because the $l$-integral already diverges independently.
BPHZ renormalization: subtracting the subdivergence
$$R\,\Sigma_{\text{nest}} = \Sigma_{\text{nest}} + \Sigma_{\text{ct-insert}} - \delta_{\text{overall}}(\Sigma_{\text{nest}} + \Sigma_{\text{ct-insert}})$$
Zimmermann's forest formula instructs us to handle divergences from the inside out. First, subtract the subdivergence by adding a counterterm-insertion diagram (next step). The combination $\Sigma_{\text{nest}} + \Sigma_{\text{ct-insert}}$ is free of subdivergences but still has an overall $1/\epsilon$ pole from the remaining $k$-integration. A final overall subtraction removes this, leaving a finite renormalized result. The nested diagram has four forests: $\varnothing$ (the empty forest, corresponding to the unsubtracted diagram), $\{\gamma_{\text{inner}}\}$ (the subdivergence alone), $\{\Gamma\}$ (the overall diagram alone), and $\{\gamma_{\text{inner}}, \Gamma\}$ (the subdivergence plus the full diagram). The forest formula sums over all four, and Zimmermann's theorem guarantees the result is finite.
The counterterm-insertion diagram
$$-i\Sigma_{\text{ct-insert}}(p) = -e^2\int\frac{d^dk}{(2\pi)^d}\;\gamma^\mu\,\frac{\not{p}-\not{k}+m}{(p-k)^2-m^2}\;\Big[-i\,\delta\Sigma^{(1)}(p-k)\Big]\;\frac{\not{p}-\not{k}+m}{(p-k)^2-m^2}\;\gamma_\mu\;\frac{1}{k^2}$$
Replace the inner one-loop self-energy subdiagram with its counterterm $\delta\Sigma^{(1)} = -(\delta_1 Z_2\,(\not{p}-\not{k}) - \delta_1 m)$, where $\delta_1 Z_2$ and $\delta_1 m$ are the one-loop renormalization constants. This is a one-loop diagram (only the $k$-integration remains) with a counterterm vertex $\otimes$ inserted on the fermion line between $v_2$ and $v_3$. By construction, adding this diagram to $\Sigma_{\text{nest}}$ cancels the $1/\epsilon$ pole from the $l$-integration — the subdivergence is gone. What remains is a one-loop integral in $k$ with at most a single overall $1/\epsilon$ divergence.
The overall divergence and final subtraction
$$\Sigma_{\text{nest}}(p) + \Sigma_{\text{ct-insert}}(p) = \frac{\alpha^2}{(4\pi)^2}\left[\frac{A_2\,m + B_2\,\not{p}}{\epsilon} + \text{finite}\right]$$
After combining the bare nested diagram with the counterterm-insertion diagram, the result has no $1/\epsilon^2$ poles (those cancelled between the two diagrams) but retains a single $1/\epsilon$ pole from the overall $k$-integration. This overall divergence is absorbed by the two-loop contributions to the mass and field-strength counterterms $\delta_2 m$ and $\delta_2 Z_2$. The cancellation of the double pole $1/\epsilon^2$ is a non-trivial consistency check: it confirms that the subdivergence was correctly identified and subtracted.
The renormalized contribution
$$\Sigma_{\text{nest}}^{\text{ren}}(p) = \frac{\alpha^2}{(4\pi)^2}\Big[c_1\,m + c_2\,\not{p} + c_3\,m\ln\!\frac{m^2}{\mu^2} + c_4\,\not{p}\ln\!\frac{m^2}{\mu^2} + c_5\,m\ln^2\!\frac{m^2}{\mu^2} + \cdots\Big]$$
The fully renormalized nested self-energy is finite and scheme-dependent. Its structure is fixed by Lorentz covariance: $\Sigma^{\text{ren}} = A(p^2) + B(p^2)\not{p}$, with $A$ and $B$ containing $\ln(m^2/\mu^2)$ terms from the renormalization scale $\mu$. The appearance of $\ln^2$ terms (double logarithms) is characteristic of two-loop diagrams with nested subdivergences — they arise from the interplay between the subtracted subdivergence and the overall integration. This diagram, together with the overlapping and vacuum-polarization-insertion topologies, makes up the full $O(\alpha^2)$ correction to the electron propagator.