Do open embeddings preserve the intersection pairing? In the following, all (co-)homology is taken with $\mathbb{Z}$-coefficients. Let $M$ be a smooth, oriented, $n$-dimensional manifold. Then, there is a Poincaré duality isomorphism $D\colon H_c^{n-k}(M)\rightarrow H_k(M)$ for all $k$. This induces the intersection pairing
$$\langle-,-\rangle_M\colon H_k(M)\times H_l(M)\stackrel{D^{-1}\times D^{-1}}{\longrightarrow}H_c^{n-k}(M)\times H_c^{n-l}(M)\stackrel{\cup}{\longrightarrow}H_c^{2n-(k+l)}(M)\stackrel{D}{\longrightarrow}H_{k+l-n}(M).$$
Now, assume $K,L\subseteq M$ are smooth, oriented submanifolds of dimension $k$ and $l$ respectively, which intersect transversely. Then, $K\cap L\subseteq M$ is a smooth, oriented submanifold of dimension $k+l-n$. The orientation here is determined by the short exact sequence $0\rightarrow T_p(K\cap L)\rightarrow T_pK\oplus T_pL\rightarrow T_pM\rightarrow0$ for all $p\in K\cap L$. The geometric nature of the intersection pairing is the observation that $\langle(i_K^M)_{\ast}[K],(i_L^M)_{\ast}[L]\rangle_M=(i_{K\cap L}^M)_{\ast}[K\cap L]$, where the $i$'s denote the respective inclusions.
Now, let $N$ be another smooth, oriented manifold and $f\colon M\rightarrow N$ an open, orientation-preserving embedding (necessarily, $M$ and $N$ have the same dimension). If $K,L\subseteq M$ are as before, then $f(K),f(L)\subseteq N$ are also smooth, oriented submanifolds, since $f$ is an embedding, whose intersection remains transverse, since $f$ is injective and induces an isomorphism on each tangent space. Since $f$ is orientation-preserving, it follows that the induced orientation on $f(K)\cap f(L)$ is the same as the one transported via $f$ from $K\cap L$. Thus, on homology, we obtain $f_{\ast}(i_K^M)_{\ast}[K]=(i_{f(K)}^N)_{\ast}[f(K)]$, $f_{\ast}(i_L^M)_{\ast}[L]=(i_{f(L)}^N)_{\ast}[f(L)]$ and $f_{\ast}(i_{K\cap L}^M)_{\ast}[K\cap L]=(i_{f(K\cap L)}^N)_{\ast}[f(K\cap L)]=(i_{f(K)\cap f(L)}^N)_{\ast}[f(K)\cap f(L)]$. This implies that $f_{\ast}\langle(i_K^M)_{\ast}[K],(i_L^M)_{\ast}[L]\rangle_M=\langle f_{\ast}(i_K^M)_{\ast}[K]f_{\ast}(i_L^M)_{\ast}[L]\rangle_N$, i.e. the induced map on homology preserves the intersection pairing of homology classes that can be represented by the inclusions of smooth, oriented submanifolds.
This is a very straightforward geometric argument, but the algebraic situation does not seem so clear at all. If $f$ were a proper map, it would induce maps $f^{\ast}\colon H_c^k(N)\rightarrow H_c^k(M)$ on compactly supported cohomology  for all $k$, and naturality of the cap product (which the Poincaré duality isomorphism is given by) together with the fact that an open embedding necessarily preserves relative fundamental classes in the appropriate sense would yield a commutative diagram
$$\require{AMScd}
\begin{CD}
H_k(M)\times H_l(M) @>{D^{-1}\times D^{-1}}>> H_c^{n-k}(M)\times H_c^{n-l}(M) @>{\cup}>> H_c^{2n-(k+l)}(M) @>{D}>> H_{k+l-n}(M)\\
@V{f_{\ast}}VV @A{f^{\ast}}AA @A{f^{\ast}}AA @V{f_{\ast}}VV  \\
H_k(N)\times H_l(N) @>{D^{-1}\times D^{-1}}>> H_c^{n-k}(N)\times H_c^{n-l}(N) @>{\cup}>> H_c^{2n-(k+l)}(N) @>{D}>> H_{k+l-n}(N)
\end{CD}.$$
Thus, $f_{\ast}$ would preserve the intersection pairing. However, an open embedding is only proper if it's the inclusion of a connected component, which is a rather unexciting case. This begs the following question: Do open embeddings in general preserve the intersection pairing?
 A: I believe I just realized the answer is yes. The crux is that the Poincaré dual of a chain can be chosen to be supported in any arbitrarily small neighborhood supporting the chain. Thus, the Poincaré dual of an element in the image of $f_{\ast}$ can be chosen to be supported in the open subset $U=f(M)\subseteq N$, but then the preimage of this support under $f$ remains compact. Thus, while $f$ not being proper means that we don't get an induced map and a diagram like in the question post, we can mimic the diagram in terms of elements.
Lemma 1: Let $M$ be an oriented manifold, $U\subseteq M$ an open subset and $K\subseteq U$ be compact. Then, $U$ inherits the structure of an oriented manifold from $M$. If $j\colon U\rightarrow M$ is the inclusion, the fundamental classes obey the relationship $j_{\ast}[U,U-K]=[M,M-K]$.
Proof: Let $n=\dim(M)$. The diagram
$$\require{AMScd}
\begin{CD}
H_n(U,U-K) @>>> H_n(M,M-K)\\
@VVV @VVV\\
H_n(U,U-x) @>>> H_n(M,M-x)
\end{CD},$$
where each map is induced by inclusion, is commutative for each $x\in K$. The bottom map is an isomorphism identifying the local orientations by definition of the induced orientation of $U$. The claim follows.$\square$
Lemma 2: Let $M$ be an oriented manifold, $U\subseteq M$ be open and $j\colon U\rightarrow M$ the inclusion. If $\beta\in H_k(U)$, $K\subseteq U$ is compact and $\psi\in H^{n-k}(U,U-K)$ represents a Poincaré dual of $\beta$, then $(j^{\ast})^{-1}\psi\in H^{n-k}(M,M-K)$ represents a Poincaré dual of $j_{\ast}\beta\in H_k(M)$.
Proof: Since $U$ is an open subspace of $M$, it inherits the structure of an oriented manifold from $M$, so Poincaré duality on $U$ makes sense. Excision implies that $j^{\ast}\colon H^{n-k}(M,M-K)\rightarrow H^{n-k}(U,U-K)$ is an isomorphism, so the statement makes sense. Using Lemma 1, we calculate
$$j_{\ast}\beta=j_{\ast}(j^{\ast}(j^{\ast})^{-1}\psi\cap[U,U-K])=(j^{\ast})^{-1}\psi\cap j_{\ast}[U,U-K]=(j^{\ast})^{-1}\psi\cap[M,M-K].$$
The claim follows.$\square$
Theorem: Let $f\colon M\rightarrow N$ be an orientation-preserving, open embedding between two oriented manifolds. Then $f$ preserves the intersection pairing.
Proof: Let's write $U=f(M)\subseteq N$, $j\colon U\rightarrow N$ for the inclusion and $\overline{f}\colon M\rightarrow U$ for the homeomorphism obtained by corestricting $f$, such that $j\overline{f}=f$. Let $\alpha\in H_k(M),\beta\in H_l(M)$ be arbitrary. Choose compact $K\subseteq U$ (resp. $L\subseteq U$) and $\varphi\in H^{n-k}(U,U-K)$ (resp. $\psi\in H^{n-l}(U,U-L)$), such that $\varphi$ (resp. $\psi$) represents a Poincaré dual of $\overline{f}_{\ast}\alpha$ (resp. $\overline{f}_{\ast}\beta$) by Poincaré duality. Now, because $K\subseteq U$ and $\overline{f}$ is a homeomorphism, $f^{-1}(K)=\overline{f}^{-1}(K)\subseteq M$ is compact. The cohomology class $\overline{f}^{\ast}\varphi\in H^{n-k}(M,M-f^{-1}(K))$ satisfies
$$\overline{f}_{\ast}(\overline{f}^{\ast}\varphi\cap[M,M-f^{-1}(K)])=\varphi\cap\overline{f}_{\ast}[M,M-f^{-1}(K)]=\varphi\cap[U,U-K]=\overline{f}_{\ast}\alpha.$$
Since $\overline{f}$ is a homeomorphism, $\overline{f}^{\ast}\varphi\cap[M,M-f^{-1}(K)]=\alpha$. Thus, $\overline{f}^{\ast}\varphi$ represents the Poincaré dual of $\alpha$. Analogously, $\overline{f}^{\ast}\psi$ represents the Poincaré dual of $\beta$. On the other hand, $(j^{\ast})^{-1}\varphi$ (resp. $(j^{\ast})^{-1}\psi$) represents the Poincaré dual of $j_{\ast}\overline{f}_{\ast}\alpha=f_{\ast}\alpha$ (resp. $j_{\ast}\overline{f}_{\ast}\beta=f_{\ast}\beta$) by Lemma 2. Thus, $$\langle f_{\ast}\alpha,f_{\ast}\beta\rangle_N=((j^{\ast})^{-1}\varphi\cup(j^{\ast})^{-1}\psi)\cap[N,N-K\cap L]=(j^{\ast})^{-1}(\varphi\cup\psi)\cap[N,N-K\cap L].$$
On the other hand,
$$\langle\alpha,\beta\rangle_M=(\overline{f}^{\ast}\varphi\cup\overline{f}^{\ast}\psi)\cap[M,M-f^{-1}(K)\cap f^{-1}(L)]=\overline{f}^{\ast}(\varphi\cup\psi)\cap[M,M-f^{-1}(K\cap L)].$$
Putting everything together,
\begin{align*}
f_{\ast}\langle\alpha,\beta\rangle_M&=j_{\ast}\overline{f}_{\ast}(\overline{f}^{\ast}(\varphi\cup\psi)\cap[M,M-f^{-1}(K\cap L)])\\
&=j_{\ast}((\varphi\cup\psi)\cap\overline{f}_{\ast}[M,M-f^{-1}(K\cap L)])\\
&=j_{\ast}((\varphi\cup\psi)\cap[U,U-K\cap L])\\
&=j_{\ast}(j^{\ast}(j^{\ast})^{-1}(\varphi\cup\psi)\cap[U,U-K\cap L])\\
&=(j^{\ast})^{-1}(\varphi\cup\psi)\cap j_{\ast}[U,U-K\cap L]\\
&=(j^{\ast})^{-1}(\varphi\cup\psi)\cap[N,N-K\cap L]\\
&=\langle f_{\ast}\alpha,f_{\ast}\beta\rangle_N.\end{align*}
