I'm going to make this more geometric and less algebraic. But it all translates to the algebro-geometric setting of divisors if you wish.
You should think of a line bundle as a twisted product, and tensor product means that you concatenate or superimpose the twists. For example, thinking of the Möbius strip as a real line bundle $\mathscr L \to S^1$, then $\mathscr L\otimes\mathscr L$ corresponds to the Möbius strip with two half-twists. You can think of complex line bundles analogously.
To make this rigorous, it's best to think about the transition function construction of a line bundle. We take an appropriate open cover $U_i$ over which the line bundles $\mathscr L$ and $\mathscr L'$ are trivial, and take transition functions $\phi_{ij}$ and $\phi'_{ij}$ on $U_i\cap U_j$. Then the tensor product has transition functions $\phi_{ij}\phi'_{ij}$ (which then concatenates or superimposes the respective twists).
It now follows that the dual or inverse line bundle $\mathscr L^*$ has transition functions $1/\phi_{ij}$. That is, it twists precisely the opposite of the original bundle, and when you tensor the two bundles you have no twist at all.
(By the way, this intuitive geometric notion of twist aligns with the tensor product notion in commutative algebra as giving extension of scalars.)
We need a more local definition of orientability to answer your question. One way to do this is to say that for any point $p$ on an $n$-manifold $M$, a local orientation at $p$ is choice of a generator $g_p$ of the relative homology group $H_n(M, M \setminus p)$ (which is isomorphic to $\Bbb Z$ by excision). A global orientation on $M$ is then choice of an orientation at $x$ for every $x\in M$ so that the choice is "consistent", in the sense that for any point $p \in M$ there is a chart around $p$ containing a ball $B$ of finite radius such that all the orientations $g_x$ for $x \in B$ are images of one single generator $g_B$ of $H_n(M, M \setminus B)$ by the isomorphism $H_n(M, M \setminus B) \to H_n(M, M \setminus x)$ induced from the inclusion $(M, M \setminus B) \hookrightarrow (M, M \setminus x)$.
There's a curious construction you could do using this machinery. Namely, consider the set $\widetilde{M}$ of all local orientations at all the points of $M$. There's a projection map $f: \widetilde{M} \to M$ that sends each local orientation to the point it orients, i.e., $f(g_p) = p$. Clearly every fiber of $f$ has cardinality two, because there are exactly two local orientations possible at each point on the manifold ($\pm 1$ are the only possible generators of $\Bbb Z$). Give $\widetilde{M}$ the topology generated by the basis of sets of the form $\mathcal{U}(g_B)$ consisting of orientations $g_x$ which are images of the generator $g_B$ of $H_n(M, M \setminus B)$ by the map $H_n(M, M \setminus B) \to H_n(M, M \setminus x)$ for balls $B$ of finite radius on $M$. This makes $f$ into a two fold covering map. $\widetilde{M}$ is known as the "orientation double cover"
Notice that local orientations of $M$ at a point $x$ are exactly same as a fiber $f^{-1}(x)$ of this covering projection. A global orientation is a section/trivialization of the orientation double cover. There's a natural morphism $H_n(M) \to \Gamma$ where $\Gamma$ is the $\Bbb Z$-module generated by the global sections of the orientation double cover; this is given by sending each homology class $\alpha \in H_n(M)$ to the "section" $s(x) = \alpha_x$ where $\alpha_x$ is image of $\alpha$ by the homomorphism $H_n(M) \to H_n(M, M \setminus x)$. $s$ is not a section of the orientation cover because $\alpha_x$ is not necessarily a generator of $H_n(M, M \setminus x)$, but it is a multiple of a section of the orientation cover. This is in fact an isomorphism (Hatcher Chapter 3.3., Lemma 3.27).
If $M$ is not orientable, $\widetilde{M}$ does not admit a global section ($\Gamma \cong 0$) which immediately implies $H_n(M) = 0$ (and vice versa). If it is orientable, $\widetilde{M}$ admits a global section which generates $\Gamma$. The morphism $\Gamma \to H_n(M, M \setminus x) \cong \Bbb Z$ for any $x \in M$, sending a section to it's value on the fiber $f^{-1}(x)$ is an isomorphism irrespective of the chosen $x$, as $M$ is connected. Hence, orientability of $M$ implies $H_n(M) \cong \Bbb Z$. This explains why your definition is equivalent to this one.
Suppose $M$ is a closed surface not containing embedded Moebius strips. Pick a point $p$ on $M$ and define an orientation on it by choosing a generator $g_p$ of $H_2(M,M \setminus p)$. For any other point $q$ on $M$, choose a path $\gamma$ joining $p$ and $q$, take a tubular neighborhood $U_\gamma$ of the path and consider the diagram $$H_2(M, M \setminus p) \leftarrow H_2(M, M \setminus U_\gamma) \rightarrow H_2(M, M \setminus q)$$ where both of the arrows are induced from the inclusion maps $(M, M \setminus U_\gamma) \hookrightarrow (M, M \setminus p)$ and $(M, M \setminus U_\gamma) \hookrightarrow (M, M \setminus q)$. By a long exact sequence argument, you can argue these are both isomorphisms. So push the generator of $H_2(M, M \setminus p)$ to $H_2(M, M \setminus q)$ using the sequence of arrows in this diagram. This gives an orientation $g_q$ at $q$.
To prove that this orientation is canonical we must verify that it does not depend on the path $\gamma$ chosen from $p$ to $q$. Suppose $\gamma'$ is another path from $p$ to $q$ that gives the orientation $-g_q$ at $q$ in the above process. This means the loop $\gamma' \cdot \gamma^{-1}$ based at $q$ is "orientation reversing", i.e., transports orientation from $g_q$ to $-g_q$. This means $\gamma' \cdot \gamma^{-1}$ lifts to a path on the orientation double cover $\widetilde{M}$ joining the two preimages in the fiber $f^{-1}(p)$. Taking $V = U_\gamma \cup U_{\gamma'}$ to be a neighborhood of this loop (where $U_{\gamma'}$ is a tubular neighborhood of $\gamma'$, similarly), we can say that this means $f$ restricts to a connected orientation double cover $f : f^{-1}(V) \to V$ of $V$, i.e., $V$ is non-orientable.
Surfaces are smoothable, so giving $M$ a smooth structure, we can invoke tubular neighborhood theorem to say that $V$ is topologically an interval bundle over a circle. There are only two such objects - $S^1 \times (0, 1)$ and the (open) Moebius strip, only the latter of which admits a connected orientation double cover. This means $V$ is homeomorphic to a Moebius strip, contradicting hypothesis on $M$. Thus $g_q$ does not depend on the chosen path $\gamma$, and we could use the technique to canonically push the chosen orientation $g_p$ to an orientation $g_x$ on every point $x \in M$ by a path to have a consistent global orientation on $M$. Thus, $M$ is orientable.
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The following code produces roughly what you're looking for: