In the proof of Lemma 3.5 in 3-Manifold Topology, the commutativity (up to a sign) of the following diagram is utilized:
How to show the commutativity?
In fact the diagram 
commutes up to a sign, where $M$ is a compact manifold.(Exercise 3.3.34 of Algebraic Topology)
The vertical isomorphisms are given by Poincare duality. For $\partial M$ this is the given by the cap product with the fundamental class $[\partial M]\in H_2\partial M$
$[\partial M]\cap (-):H^k\partial M\rightarrow H_{2-k}\partial M$.
For the relative case this is induced by the cap product with the fundamental class $[M,\partial M]\in H_3(M,\partial M)$. There are two of these we need to use
$[M,\partial M]\cap(-):H^kM\rightarrow H_{3-k}(M,\partial M)$
$[M,\partial M]\cap(-):H^k(M,\partial M)\rightarrow H_{3-k}M$.
The cap product is discussed at the beginning of "The Duality Theorem" in section 3.3 of Hatcher's Algebraic Topology.
The cap product satisfies a certain naturality, meaning that it is compatible with long exact sequences, and in particular with their boundary homomorphisms. This is discussed briefly on pages 240-241 (at least in the updated copy that I have). Hatcher gives the formula
$f_*(\alpha)\cap\varphi=f_*(\alpha\cap f^*(\varphi))$
for a map (possibly of pairs) $f:(X,A)\rightarrow (Y,B)$. There is some interpretation to be made in this brief discussion, and I'll refer you to Tammo tom Dieck's "Algebraic Topology" section 18 for a fuller discussion. I'm not sure exactly where in Hatcher it is proved.
In our case we take $f=(i,j):(\partial M,\emptyset)\hookrightarrow (M,\partial M)$ to be the inclusion of pairs. Referring back to Hatcher, pg. 240, there is given a formula for the boundary operator on singular homology
$\partial(\sigma\cap\varphi)=(-1)^l(\partial\sigma\cap\varphi-\sigma\cap\delta\varphi)$
for $\sigma\in S_k(X,A)$ and $\varphi\in S^l(A)$ (this is the boundary operator $S_l(X)\rightarrow S_{l-1}(X)$, not the connecting homomorphism of the pair $(X,A)$).
Consider now the definition of the connecting homomorphism $\Delta:H_l(M,\partial M)\rightarrow H_{l-1}(\partial M)$. If $\partial\sigma=i_*\rho$ for $\rho\in S_{l-1}(\partial M)$, then $\Delta([\sigma])=[\rho]$ (the square brackets here denote homology classes). We have
$\partial(\sigma\cap\varphi)=(-1)^l(\partial\sigma\cap\varphi-\sigma\cap\delta\varphi)=(-1)^li_*(\rho)\cap\varphi=(-1)^li_*(\rho\cap i^*\varphi)$
where we have assumed that $\varphi$ is a cocycle, and used the naturality formula given previously. We conclude from this that for $[\sigma]\in H_k(M,\partial M)$ and $[\varphi]\in H^l(\partial M)$ we have $[\sigma\cap\varphi]\in H_{k-l}(M,\partial M)$ and
$\Delta[\sigma\cap\varphi]=(-1)^l[\rho\cap i^*\varphi]\in H_{k-l-1}(\partial M).$
Taking $[\sigma]=[M,\partial M]\in H_3(M,\partial M)$ and $\varphi\in H^1(\partial M)$, we note that $\Delta[M,\partial M]=[\partial M]$ (exercise), so that $[\rho]=[\partial M]$ and the formula above becomes
$\Delta([M,\partial M]\cap\varphi)=-[\partial M]\cap i^*\varphi\in H_1(\partial M)$.
In light of the definitions given previously for Poincare duality we see that this is just the left-most square of your first diagram, which thus commutes up to sign.
Signed commutativity of the right-hand square is proved similarly, and I'll leave this for you to check. Please let me know if anything I've written is confusing, or if you have any trouble with the second square.