Are you sure this is true?
Consider the following: take $M = S^1$ to be the unit circle and take $B \to M$ to be trivial line bundle with fiber $\mathbb R^1$. The space of smooth global sections of $B$ is then just the space of smooth functions on $S^1$. Equip $S^1$ with the usual metric on the circle and equip $B$ with the Euclidean metric. Take the connections involved to be the trivial ones.
Now let $f : S^1 \to [-1,1]$ be the function induced by $x \mapsto \sin(2\pi x)$. This function is smooth, and thus a smooth section of $B$. Note that the absolute value function $|\cdot| : [-1,1] \to \mathbb R$ is continuous. By the Weierstrass approximation theorem there exists a sequence of polynomials $p_m$ that converges uniformely to $|\cdot|$ on $[-1,1]$. The composition $s_m = p_m \circ f$ is thus a sequence of smooth sections of $B$, that converges to $s(x) = |\sin(2\pi x)|$, which is not a smooth section of $B$ (its derivative is discontinuous at the point corresponding to the zero $\pi$ - just plot the graph of the function to see it).
I haven't verified the details, so it is possible that the sequence $(s_m)$ does not converge in the seminorm required (basically due to the discontinuity of the first derivative of the limit). If so, then replacing the absolute value by a function with continuous first, but not second, derivatives should give a counterexample.
In general I believe one obtains the space of $L^2$ sections of $B$ by taking the completion with respect to the seminorm you defined, and not the space of smooth sections.
Note that the formal definition of the pullback of $\pi : L \to Y$ by $f : X \to Y$ is as follows: the total space is given by $f^*L = \{(x, \ell) \in X\times L \mid f(x) = \pi(\ell)\}$ and the projection $\pi' : f^*L \to X$ is given by projection onto the first factor; i.e. $\pi'(x, \ell) = x$. There is also a map $F : f^*L \to L$ given by projection onto the second factor, i.e. $F(x, \ell) = \ell$. These maps give rise to a commutative square:
$$\require{AMScd}
\begin{CD}
f^*L @>{F}>> L\\
@V{\pi'}VV @VV{\pi}V \\
X @>{f}>> Y
\end{CD}$$
That is, $\pi\circ F = f\circ\pi'$.
Note that for $x_0 \in X$, we have $$(f^*L)_{x_0} = \pi'^{-1}(x_0) = \{(x_0, \ell) \in X\times L \mid f(x_0) = \pi(\ell)\} \cong \pi^{-1}(f(x_0)) = L_{f(x_0)}.$$
The pullback metric $f^*h$ on $(f^*L)_{x_0}$ is just the original metric $h$ on $L_{f(x_0)}$. More precisely, $$(f^*L)_x(s(x), t(x)) = h_{f(x)}(F(s(x)), F(t(x))).$$
Best Answer
Let $L \to \Sigma$ be your holomorphic line bundle. This bundle admits, by a partition of unity argument and the local trivialization of the bundle, a smooth bundle metric, lets call it $g_L$. Given a global section $s:\Sigma \to L$, we define $$ |s|^2=g_{L}(s,s). $$ This is well defined and we can just integrate it: $$ ||s||_{L^2}^2=\int_{\Sigma}|s|^2 d\mathrm{vol} $$
In more detail: So given a trivialization atlas $\{(U_i,\phi_i) \}$ of $L$, this section looks locally like $$ s:U_i \to U_i \times \mathbb{C}, z \mapsto (z,u(z)) $$ and a local metric looks like $$ g_i:U_i\times \mathbb{C}\times \mathbb{C} \to \mathbb{R},(z,u_1,u_2) \mapsto u_1 \overline{u_2}. $$ You can glue these local metrics together using a partition of unity argument to build a (global) metric $g_L$.
To see that this is well defined for your section, use the fact that for a hermitian vector bundle, the transition functions can be chose unitary, in your case: $g_{ij} \in U(1)$. But you see that every local metric is invariant under $u(1)$ transformations: $$ e^{i\phi}u_1 \overline{e^{i\phi}u_2}=u_1 \overline{u_2} $$ and so is a combination of them: The (global) metric! Now, this gives therefore a well-defined function for a section $s: \Sigma \to L$ in the form $$ |s|^2: \Sigma \to \mathbb{R}, z \mapsto g_L(s,s) . $$ This is now something you can integrate over $\Sigma$. Now, every section with finite $L^2$ norm is in $L^2(\Gamma(L))$ (the $L^2$-space of sections of $L$).
A good exercise is to check what this would look like for the space of sections of the trivial bundle $\Sigma \times \mathbb{C} \to \Sigma$ for a (non-)compact Riemann surface. Hint: It is the ordinary $L^2$ norm $$ \int_{\Sigma} |u(z)|^2 d\mathrm{vol}. $$ I just added non-compact to allow non-trivial, holomorphic sections.