1) comes down to a statement from multilinear algebra: if $V$ is a free $k$-module of rank $n$, there is a natural pairing $\wedge^n(V) \otimes \wedge^n(V^*) \to k$ given by mapping $v_1 \wedge \dotsc \wedge v_n \otimes \phi_1 \wedge \dotsc \wedge \phi_n$ to the determinant of the matrix $(\phi_i(v_j))$. It is perfect, i.e. induces a natural isomorphism $\wedge^n(V)^* \cong \wedge^n(V^*)$. You can check this directly using bases.
For the rest of the questions, you should explaind your notation ($Y,X,I$, etc.).
Your question: "For an affine open $U \subseteq X$, we have
$$\text{O1}. \text{ }\Omega^1_X(U) \cong \Omega^1_{\mathcal{O}_X(U)}.$$
For affine open sets $V\subseteq U \subseteq X$ we have
$$\text{O2}. \text{ } \Omega^1_{\mathcal{O}_X(V)} \cong \Omega^1_{\mathcal{O}_X(U)}\otimes_{\mathcal{O}_X(U)} \mathcal{O}_X(V),$$
and furthermore, the restriction map $\Omega^1_X(U) \rightarrow \Omega^1_X(V)$ is given by $ω \rightarrow ω\otimes 1$.
It seems to me the following is true:
Question O1. Since your sheaf $\Omega^1_X$ is defined as the sheafification $F^{+}$ of the pre-sheaf $F(U):= \Omega^1_{\mathcal{O}_X(U)}$, it seems to me $F^{+}$ and $F$ should have she same sections on affine open subsets $U:=Spec(A) \subseteq X$. Hence
$$F(U)=F^{+}(U) \cong \Omega^1_{A/k}:=\Omega^1_{\mathcal{O}_X(U)}.$$
Question O2. The construction of the module of differentials in functorial in the sense that for any open subscheme $U \subseteq X$ it follows $i^*(\Omega^1_{X/k}) \cong \Omega^1_{U/k}$ where $i: U \rightarrow X$ is the inclusion map.
Moreover for any pair of maps of commutative rings $A \rightarrow B$ and $A \rightarrow S$
let $B_S :=S\otimes_A B$. There is (Matsumura's book "Commutative ring theory", Exercise 25.4 page 198) a canonical isomorphism
$$ B_S \otimes_B \Omega^1_{B/A} \cong \Omega^1_{B_S/S}.$$
Hence for any inclusion of basic open sets $D(f) \subseteq D(g) \subseteq U:=Spec(A)$
you should get the following:
$$(\Omega^1_{D(g)/k})_{D(f)}:= \Omega^1_{A_g/k} \otimes_{A_g} A_f \cong \Omega^1_{A_{fg}/k}\cong \Omega^1_{A_f/k}.$$
By $(\Omega^1_{D(g)/k})_{D(f)}$ I mean the restriction of $\Omega^1_{D(g)/k}$
to the open set $D(f)$. Here we have used that $D(f)\cap D(g):=D(fg) =D(f)$ since $D(f) \subseteq D(g)$.
The restriction map
$$\rho: \Omega^1_{D(g)/k}\rightarrow (\Omega^1_{D(g)/k})_{D(f)} := \Omega^1_{D(g)/k}\otimes_{A_g} A_f\cong \Omega^1_{D(f)/k}$$
is the canonical map $\rho(\omega):=\omega \otimes 1$ where $1\in A_f$ is the multiplicative identity. Formula O2 should follow since $U$ and $V$ have open covers on the form $D(f)$.
Best Answer
By Hartshorne II 5.2, if $\alpha : A \to B$ is a ring homomorphism and $f : {\rm Spec}(B) \to {\rm Spec}(A)$ is the corresponding morphism of affine schemes, and if $M$ is an $A$-module, then $f^\star(\widetilde M) = (M\otimes_A B)^\sim$, where $B$ gets its $A$-module structure from the ring homomorphism $\alpha$. We can apply this to our setup. Here, $A = B \otimes_k B$ and $M = I/I^2$, and the morphism $f$ is the diagonal morphism $\delta$ corresponding to the ring homomorphism $\alpha : A \to B$. So $\delta^\star((I/I^2)^\sim)$ is equal to $\left( (I / I^2) \otimes_A B\right)^\sim$.
But since the ring morphism $\alpha : A \to B$ is surjective, and since $I$ is the kernel of $\alpha$, it is clear that $B$ is really the same thing as $A / I$. So $(I / I^2) \otimes_A B \cong (I/I^2) \otimes_A (A/I)$, and this is isomorphic to $I/I^2$ as an $A$-module via the isomorphism $I/I^2 \to (I/I^2)\otimes_A (A/I)$ given by $[x] \mapsto [x]\otimes_A 1$ for every $[x] \in I/I^2$. (Here, $x$ is an element of $I$, and $[x]$ denotes its equivalence class modulo $I^2$. This old post may help, if you need further convincing.)
However, to characterise $\delta^\star((I/I^2)^\sim)$ as a quasicoherent sheaf on ${\rm Spec}(B)$, we really need to think of $((I/I^2) \otimes_A B)$ as a $B$-module rather than merely as an $A$-module. So we need to understand how to multiply an element $[x] \in I/I^2$ by an element $b \in B$. Identifying $[x]$ as the element $[x]\otimes_A 1 \in (I/I^2)\otimes_A(A/I)$, the multiplication is given by $b.([x]\otimes_A 1)=[x]\otimes_A b$. Now notice that $b=\alpha(b \otimes_A 1)$, where $\alpha : A \to B$ is the diagonal homomorphism. So if $x = \sum_i b_{1,i} \otimes_A b_{2,i}$, then $[x]\otimes_A b= [(b\otimes_A 1).(\sum_{i}b_{1,i} \otimes_A b_{2,i})]\otimes_A 1=[\sum_i bb_{1,i} \otimes_A b_{2,i}]\otimes_A 1$, which is identified with the element $[\sum_i bb_{1,i} \otimes_A b_{2,i}] \in I/I^2$ via the isomorphism $I/I^2 \cong (I/I^2)\otimes_A(A/I)$. This is precisely the multiplication rule that you described.