Some definitions
Recall that given two ($i=1,2$) Hilbert $B_i$-modules $E_i$ and a $*$-homomorphism $$\phi:B_1\to \mathcal{L}(E_2),$$ where $\mathcal{L}(E)$ denotes the space of adjointable operators on the Hilbert module $E$, the balanced product between $E_1$ and $E_2$, denoted by $E_1\otimes_{\phi} E_2$, is given as follows: consider the $B_2$-valued inner product $$\langle a_1\otimes b_1,a_2\otimes b_2\rangle_{\phi}:=\langle b_1,\phi\left(\langle a_1,a_2\rangle_{E_1}\right)b_2\rangle_{E_2}. $$ Then, by letting $\mathcal{N}_{\phi}:=\{ v \in E_1\otimes E_2:\langle v,v\rangle_{\phi}=0 \}$, we can define $E_1\otimes_{\phi} E_2$ as the completion of $E_1\otimes E_2/\mathcal{N}_{\phi}$ with respect to the inner product defined above.
Now, let $X$ be a complete, connected Riemannian manifold, and let $F\to X$ be a hermitian vector bundle. Recall that the Hilbert space $L^2(X)$ is the completion of the space of compactly supported smooth sections $\Gamma^{\infty}_c(F)$, with respect to the inner product $$\langle s_1,s_2\rangle_{L^2}:=\int_X\langle s_1(x),s_2(x)\rangle_{F_x}d\mu(x), $$ where $\mu$ denotes the canonical Riemannian measure coming from the metric over $X$ and $s_1,s_2\in\Gamma^{\infty}_c(F)$. There is a $C_0(X)$-action over $L^2(X)$ given by pointwise multiplication, which we denote by $$\pi:C_0(M)\to B(L^2(X)).$$ Recall, also, that $C_0(X)$ can be perceived as a Hilbert module over itself, with $C_0(X)$-valued inner product given by $\langle f_1,f_2\rangle_{C_0(X)}:=\overline{f_1}f_2.$
The problem itself
I am trying to prove that the balanced tensor product $C_0(X)\otimes_{\pi} L^2(X)$ is unitarily isomorphic to $L^2(X)$. It is easy to see that $\mathcal{N}_{\pi}=\{0\}$, so I tried to define the following map \begin{equation*} \begin{split} F:C_0(X)&\otimes L^2(X)&&\to &&&L^2(X)\\ f&\otimes\psi && \mapsto &&& f\psi \end{split} \end{equation*} I can see this map preserves inner products, so it extends to an isometry from $C_0(X)\otimes_{\pi}L^2(X)$ to $L^2(X)$. Now, how to prove it is surjective? This is easy when $X$ is compact, since then $1\in C_0(X)$. Thus, given $\psi\in L^2(X)$, I can make $F(1\otimes\psi)=\psi$ and be done with it. My question then is: is it possible to prove this map is surjective when $X$ is non-compact? I tried to argue using approximate units, but didn't manage to get there. I guess if one can prove that $\pi$ is nondegenerate, then taking an approximate unit $\rho_n$, the operator $\pi(\rho_n)$ converges strongly to $1_{B(L^2(X))}$, which seems to be all I need.
In situations like this it is often easier to just prove that $F$ has dense image. Since it is isometric, the image is also complete, which means that it must be surjective.
In this case, you are even given a dense subspace of $L^2(X)$, namely $\Gamma^\infty_c(X)$, for which it is easy to show that it lies in the image of $F$: Let $\psi\in \Gamma^\infty_c(X)$ and $f\in C_c(X)$ such that $f=1$ on the support of $\psi$. The existence of $f$ is guaranteed by Urysohn's lemma. Clearly, $F(f\otimes \psi)=f\psi=\psi$.