$\textbf{Lemma:}$ Let $g\colon (X,A,B)\to (\overline X,\overline
A,\overline B)$ be a map between two excisive triads, then for $x\in
H^k(\overline X,\overline A), y\in H_l(X,A\cup B)$ we have $$x\frown
g_*(y)=g_*\big(g^*(x)\frown y\big)\in H_{l-k}(\overline X,\overline
B).$$

Commutativity of the red portion: The inclusion map $i\colon(U,U\backslash K)\hookrightarrow (M,M\backslash K\cap L)$ is the composition of two inclusion maps $$(U,U\backslash K)\overset{\beta}{\hookrightarrow}(U,U\backslash K\cap L)\overset{f}{\hookrightarrow}(M,M\backslash K\cap L)$$ Similarly, the inclusion map $i\colon (U\cap V,U\cap V\backslash K\cap L)\hookrightarrow(M,M\backslash K\cap L)$ is the composition of two inclusion maps $$(U\cap V,U\cap V\backslash K\cap L)\overset{\alpha}{\hookrightarrow}(U,U\backslash K\cap L)\overset{f}{\hookrightarrow}(M,M\backslash K\cap L)$$ We first show that $$H^k(M,M\backslash K\cap L)\xrightarrow{(f\circ \beta)^*}H^k(U,U\backslash K)\xrightarrow{\bullet\frown \mu_K^U} H_{n-k}(U)$$ is same as $$H^k(M,M\backslash K\cap L)\xrightarrow{(f\circ \alpha)^*}H^k(U\cap V,U\cap V\backslash K\cap L)\xrightarrow{\bullet\frown \mu_{K\cap L}^{U\cap V}}H_{n-k}(U\cap V)\xrightarrow{\alpha_*}H_{n-k}(U)$$ are the same.
So take $x\in H^k(M,M\backslash K\cap L)$ and consider $$\alpha_*\bigg((f\circ \alpha)^*(x)\frown \mu_{K\cap L}^{U\cap V}\bigg)=\alpha_*\bigg(\alpha^*\big(f^*(x)\big)\frown \mu_{K\cap L}^{U\cap V}\bigg)$$$$=f^*(x)\frown\alpha_*\left(\mu_{K\cap L}^{U\cap V}\right)=f^*(x)\frown \mu^U_{K\cap L}$$
Note that in the last equality, we are using the following uniqueness property:
For an $n$-dimensional oriented manifold $N$ and
$A\subseteq_\text{compact}N$ and for every section
$\{\varphi_x\}_{x\in A}$ we have a unique $\alpha_A\in
H_n(N,N\backslash A)$ such that inclusion induced map
$H_n(N,N\backslash A)\to H_n(N,N\backslash x)$ sends $\alpha_A$ to
$\alpha_x$ for each $x\in A$.
A similar logic gives $$\beta_*\bigg((f\circ \beta)^*(x)\frown \mu_{K}^{U}\bigg)=\beta_*\bigg(\beta^*\big(f^*(x)\big)\frown \mu_{K}^{U}\bigg)=f^*(x)\frown\beta_*\left(\mu_{K}^{U}\right)$$
Now, $\beta\colon U\hookrightarrow U$ is identity. So, $\text{Id}=\beta_*\colon H_{n-k}(U)\to H_{n-k}(U)$ hence comparing the left most and the right most sides of the above equality we have $$(f\circ \beta)^*(x)\frown \mu_{K}^{U}=f^*(x)\frown\beta_*\left(\mu_{K}^{U}\right).$$ Again, the above uniqueness property shows $$\beta_*\left(\mu_K^U\right)=\mu_{K\cap L}^U.$$ Therefore, combining all these
$$\alpha_*\bigg((f\circ \alpha)^*(x)\frown \mu_{K\cap L}^{U\cap V}\bigg)=f^*(x)\frown \mu^U_{K\cap L}$$$$=f^*(x)\frown\beta_*\left(\mu_{K}^{U}\right)$$$$=(f\circ \beta)^*(x)\frown \mu_{K}^{U}$$ We have shown the commutativity of a portion(corresponding to left summand, red shaded) of the pentagon of the above diagram.
Similarly, one can show the commutativity of the other portion(corresponding to the right summand) of the pentagon of the above diagram. The only thing left is the commutativity topmost triangle, which is easy as all maps are inclusion induced.