[proofplan]
We apply Cartan's Theorem A to the coherent analytic sheaf $\mathcal{I}_V$ on the Stein manifold $X$. The theorem gives generation of each stalk by global sections through the standard tensor product over the global holomorphic function algebra $\mathcal{O}(X)$. The only remaining point is to compare that standard Cartan map with the map stated in the theorem, whose tensor product is over $\mathbb{C}$; the latter factors through the former by the universal quotient defining tensor products over $\mathcal{O}(X)$.
[/proofplan]
[step:Fix a stalk and record the module structures used in the tensor maps]
Fix a point $p \in X$. Define the global holomorphic function algebra $\mathcal{O}(X)$ by
\begin{align*}
\mathcal{O}(X) := H^0(X,\mathcal{O}_X).
\end{align*}
Let $\gamma_p$ denote the germ homomorphism
\begin{align*}
\gamma_p: \mathcal{O}(X) &\to \mathcal{O}_{X,p} \\
h &\mapsto h_p.
\end{align*}
This makes $\mathcal{O}_{X,p}$ an $\mathcal{O}(X)$-module. The vector space $H^0(X,\mathcal{I}_V)$ is also an $\mathcal{O}(X)$-module by pointwise multiplication of a global holomorphic function with a global section of the ideal sheaf.
Define the stalk-evaluation map
\begin{align*}
\rho_p: H^0(X,\mathcal{I}_V) &\to (\mathcal{I}_V)_p \\
s &\mapsto s_p.
\end{align*}
For every $h \in \mathcal{O}(X)$ and every $s \in H^0(X,\mathcal{I}_V)$, multiplication of germs gives
\begin{align*}
\rho_p(hs) = (hs)_p = h_p s_p = \gamma_p(h)\rho_p(s).
\end{align*}
Thus $\rho_p$ is compatible with the $\mathcal{O}(X)$-module structures.
[/step]
[step:Apply Cartan's Theorem A to the coherent ideal sheaf $\mathcal{I}_V$]
The hypotheses of Cartan's Theorem A are satisfied: $X$ is a Stein manifold by assumption, and $\mathcal{I}_V$ is a coherent $\mathcal{O}_X$-module by assumption. Applying Cartan's Theorem A to the coherent sheaf $\mathcal{F} := \mathcal{I}_V$ gives, for the fixed point $p$, a surjective $\mathcal{O}_{X,p}$-linear map
\begin{align*}
\Phi_p: H^0(X,\mathcal{I}_V) \otimes_{\mathcal{O}(X)} \mathcal{O}_{X,p} &\to (\mathcal{I}_V)_p \\
s \otimes_{\mathcal{O}(X)} a &\mapsto a s_p.
\end{align*}
[guided]
We now use the only substantial theorem in the proof: Cartan's Theorem A. The theorem applies to a coherent $\mathcal{O}_X$-module on a Stein manifold and concludes that its stalks are generated by global sections. In the present setting, the sheaf to which we apply the theorem is
\begin{align*}
\mathcal{F} := \mathcal{I}_V.
\end{align*}
The required hypotheses are exactly the hypotheses of the theorem being proved: $X$ is Stein, and $\mathcal{I}_V$ is coherent as an $\mathcal{O}_X$-module. Therefore Cartan's Theorem A gives, at the fixed point $p \in X$, a surjective $\mathcal{O}_{X,p}$-linear map
\begin{align*}
\Phi_p: H^0(X,\mathcal{I}_V) \otimes_{\mathcal{O}(X)} \mathcal{O}_{X,p} &\to (\mathcal{I}_V)_p \\
s \otimes_{\mathcal{O}(X)} a &\mapsto a s_p.
\end{align*}
The tensor product here is over $\mathcal{O}(X)$ because global sections of $\mathcal{I}_V$ form an $\mathcal{O}(X)$-module, and the stalk $\mathcal{O}_{X,p}$ is an $\mathcal{O}(X)$-module through the germ homomorphism $\gamma_p: \mathcal{O}(X) \to \mathcal{O}_{X,p}$. This is the standard Cartan Theorem A generation map.
[/guided]
[/step]
[step:Factor the stated $\mathbb{C}$-tensor map through the Cartan map]
Define the $\mathbb{C}$-bilinear pairing
\begin{align*}
B_p: H^0(X,\mathcal{I}_V) \times \mathcal{O}_{X,p} &\to (\mathcal{I}_V)_p \\
(s,a) &\mapsto a s_p.
\end{align*}
By the universal property of the tensor product over $\mathbb{C}$, $B_p$ induces the $\mathbb{C}$-linear map
\begin{align*}
\Psi_p: H^0(X,\mathcal{I}_V) \otimes_{\mathbb{C}} \mathcal{O}_{X,p} &\to (\mathcal{I}_V)_p \\
s \otimes_{\mathbb{C}} a &\mapsto a s_p.
\end{align*}
This is the map appearing in the theorem statement.
Let $\pi_p$ be the canonical quotient map
\begin{align*}
\pi_p: H^0(X,\mathcal{I}_V) \otimes_{\mathbb{C}} \mathcal{O}_{X,p} &\to H^0(X,\mathcal{I}_V) \otimes_{\mathcal{O}(X)} \mathcal{O}_{X,p} \\
s \otimes_{\mathbb{C}} a &\mapsto s \otimes_{\mathcal{O}(X)} a.
\end{align*}
The relative tensor product over $\mathcal{O}(X)$ is obtained from the tensor product over $\mathbb{C}$ by imposing the balancing relations
\begin{align*}
(hs) \otimes_{\mathbb{C}} a = s \otimes_{\mathbb{C}} \gamma_p(h)a
\end{align*}
for $h \in \mathcal{O}(X)$, $s \in H^0(X,\mathcal{I}_V)$, and $a \in \mathcal{O}_{X,p}$. Hence $\pi_p$ is surjective.
For every pure tensor $s \otimes_{\mathbb{C}} a$, we have
\begin{align*}
(\Phi_p \circ \pi_p)(s \otimes_{\mathbb{C}} a)
= \Phi_p(s \otimes_{\mathcal{O}(X)} a)
= a s_p
= \Psi_p(s \otimes_{\mathbb{C}} a).
\end{align*}
By $\mathbb{C}$-linearity, $\Psi_p = \Phi_p \circ \pi_p$. Since both $\Phi_p$ and $\pi_p$ are surjective, $\Psi_p$ is surjective.
[guided]
The theorem statement uses a tensor product over $\mathbb{C}$, while Cartan's Theorem A gives a tensor product over $\mathcal{O}(X)$. We must explain why the stronger-looking $\mathbb{C}$-tensor map is also surjective.
First define the pairing whose tensor map is being discussed:
\begin{align*}
B_p: H^0(X,\mathcal{I}_V) \times \mathcal{O}_{X,p} &\to (\mathcal{I}_V)_p \\
(s,a) &\mapsto a s_p.
\end{align*}
This pairing is $\mathbb{C}$-bilinear because multiplication in the stalk $(\mathcal{I}_V)_p$ is $\mathbb{C}$-bilinear. Therefore the universal property of the tensor product over $\mathbb{C}$ gives a unique $\mathbb{C}$-linear map
\begin{align*}
\Psi_p: H^0(X,\mathcal{I}_V) \otimes_{\mathbb{C}} \mathcal{O}_{X,p} &\to (\mathcal{I}_V)_p \\
s \otimes_{\mathbb{C}} a &\mapsto a s_p.
\end{align*}
This is precisely the map written in the theorem statement.
Next compare it with the Cartan map. The tensor product over $\mathcal{O}(X)$ is a quotient of the tensor product over $\mathbb{C}$: one imposes the balancing relation saying that multiplying the global section first by $h \in \mathcal{O}(X)$ is the same as multiplying the stalk coefficient by the germ $\gamma_p(h) \in \mathcal{O}_{X,p}$. Concretely, define
\begin{align*}
\pi_p: H^0(X,\mathcal{I}_V) \otimes_{\mathbb{C}} \mathcal{O}_{X,p} &\to H^0(X,\mathcal{I}_V) \otimes_{\mathcal{O}(X)} \mathcal{O}_{X,p} \\
s \otimes_{\mathbb{C}} a &\mapsto s \otimes_{\mathcal{O}(X)} a.
\end{align*}
Because the target is formed by quotienting by the relations
\begin{align*}
(hs) \otimes_{\mathbb{C}} a = s \otimes_{\mathbb{C}} \gamma_p(h)a,
\end{align*}
the map $\pi_p$ is surjective.
Now compute the composite with the Cartan map on a pure tensor:
\begin{align*}
(\Phi_p \circ \pi_p)(s \otimes_{\mathbb{C}} a)
= \Phi_p(s \otimes_{\mathcal{O}(X)} a)
= a s_p
= \Psi_p(s \otimes_{\mathbb{C}} a).
\end{align*}
Since pure tensors span the algebraic tensor product, linearity gives
\begin{align*}
\Psi_p = \Phi_p \circ \pi_p.
\end{align*}
Cartan's Theorem A gives that $\Phi_p$ is surjective, and the quotient map $\pi_p$ is surjective by construction. Therefore their composite $\Psi_p$ is surjective.
[/guided]
[/step]
[step:Translate surjectivity into generation of the stalk by global sections]
Let $\eta \in (\mathcal{I}_V)_p$. Since $\Psi_p$ is surjective, there exists an element $\xi \in H^0(X,\mathcal{I}_V) \otimes_{\mathbb{C}} \mathcal{O}_{X,p}$ such that
\begin{align*}
\Psi_p(\xi) = \eta.
\end{align*}
By the definition of the algebraic tensor product, $\xi$ is a finite sum of pure tensors, so there exist finitely many sections $s_1,\ldots,s_m \in H^0(X,\mathcal{I}_V)$ and finitely many germs $a_1,\ldots,a_m \in \mathcal{O}_{X,p}$ such that
\begin{align*}
\xi = \sum_{j=1}^{m} s_j \otimes_{\mathbb{C}} a_j.
\end{align*}
Applying $\Psi_p$ gives
\begin{align*}
\eta
= \Psi_p(\xi)
= \sum_{j=1}^{m} a_j (s_j)_p.
\end{align*}
Thus every element of $(\mathcal{I}_V)_p$ is an $\mathcal{O}_{X,p}$-linear combination of germs at $p$ of global sections of $\mathcal{I}_V$. Since $p \in X$ was arbitrary, this holds for every point $p \in X$, and the stated map is surjective for every $p$.
[/step]
