I encounter some difficulties to understand some notations in the proof of Theorem 1.2.17 of the book "Positivity in Algebraic Geometry I" by Robert Lazarsfeld. To make things clear, a scheme is a seperated algebraic scheme of finite type over $\mathbb C$.
Theorem 1.2.17. (Amplitude in families) : Let $f : X \to T$ be a proper morphism of schemes, and $L$ a line bundle on $X$. Given $t \in T$, write $$X_t = f^{-1}(t), ~~L_t = L \vert_{X_t}$$ Assume that $L_0$ is ample on $X_0$ for some point $0 \in T$. Then there is an open neighborhood $U$ of $0$ in $T$ such that $L_t$ is ample on $X_t$ for every $t \in U$.
The proof is essentially divided into three parts and I am stuck on a notation intervening in the second one.
First step (I have no problem with this one) : we show that for any coherent sheaf $\mathcal F$ on $X$, there is a positive integer $m(\mathcal F,L)$ such that $R^i f_* \left( \mathcal F \otimes L^{\otimes m} \right) = 0$ in a neighborhood $U_m \subset T$ of $0$ (*).
Second step : we show that the canonical mapping $\rho_m : f^* f_* L^{\otimes m} \to L^{\otimes m}$ is surjective along $X_t$ for all $t$ in a neighborhood $U^{"}_m$ of $0$ provided that $m$ is sufficiently large.
To see this, apply (*) to the ideal sheaf $\mathcal I_{X_0/X}$ of $X_0$ in $X$. One finds that $$f_* \left( L^{\otimes m} \right) \to f_* \left( L^{\otimes m} \otimes \mathcal O_{X_0} \right) = H^0\left(X_0, L_0^{\otimes m} \right) (**)$$ is surjective when $m \gg 0$. But all sufficiently large powers of the ample line bundle $L_0$ are globally generated. Composing (**) with the evaluation $H^0\left(X_0, L_0^{\otimes m} \right) \otimes \mathcal O_{X_0} \to L_0^{\otimes m}$, then, shows that $\rho_m$ is surjective along $X_0$ for $m \gg 0$. (...)
I’m not sure to understand what the equality $f_* \left( L^{\otimes m} \otimes \mathcal O_{X_0} \right) = H^0\left(X_0, L_0^{\otimes m} \right)$ means. I know that by the projection formula, we have $f_* \left( L^{\otimes m} \otimes \mathcal O_{X_0} \right) = (f\vert_ {X_0})_*\left(L_0^{\otimes m} \right)$ and that this sheaf is the one associated to the presheaf
$$U \mapsto H^0\left( f^{-1}(U) \cap X_0, L_0^{\otimes m} \right)$$
but I am really confused by the notation used. Especially since, subsequently, we have to compose with $$H^0\left(X_0, L_0^{\otimes m} \right) \otimes \mathcal O_{X_0} \to L_0^{\otimes m}$$ , I am all the more lost. Do you have any indications ? Here is the proof extracted from the book for more clarity.
We're assuming $T = \operatorname{Spec} A$ is affine so $f_*(L^n \otimes \mathcal{O}_{X_0})$ is then the coherent sheaf associated to the $A$-module $H^0(X, L^n \otimes \mathcal{O}_{X_0})$, which as you observed is naturally isomorphic to $H^0(X_0, L_0^n)$ by the projection formula(or, if you'd prefer, using this theorem from the stacks project).
As such it would probably be more accurate to write $f_*(L^n \otimes \mathcal{O}_{X_0}) \cong \widetilde{H^0(X_0, L_0^n)}$, but usually forgetting the tilde is more or less harmless since the categories of quasicoherent sheaves on $T$ and the category of $A$-modules can be naturally identified with one another. Similarly, $f_*(L^n)$ is the coherent sheaf $\widetilde{H^0(X, L^n)}$ and this map $f_*(L^n) \to f_*(L^n \otimes \mathcal{O}_{X_0})$ is induced by the map of $A$-modules $H^0(X, L^n) \to H^0(X_0, L_0^n)$ obtained by restricting sections, which we find to be surjective using the first step.
Hence $H^0(X, L^n) \otimes_\mathbb{C} \mathcal{O}_{X_0} \to H^0(X_0, L_0^n) \otimes_{\mathbb{C}} \mathcal{O}_{X_0}$ is surjective, where this tensor product means the free $\mathcal{O}_{X_0}$-module generated by a $\mathbb{C}$-basis of the given vector space. By global generation, the natural restriction map $H^0(X_0, L^n_0) \otimes_\mathbb{C} \mathcal{O}_{X_0} \to L_0$ is surjective too. Hence, the composite restriction map $$H^0(X, L^n) \otimes_{\mathbb{C}} \mathcal{O}_{X_0} \longrightarrow L_0^n$$ is as well.
By the argument in the book, to show $\rho: f^*f_* L^n \to L^n$ is surjective in a neighborhood of $X_0$ it suffices to show that it's surjective when we restrict to $X_0$.
Once you convince yourself that $\rho|_{X_0}: (f^*f_*L^n)|_{X_0} \to (L^n)|_{X_0}$ is naturally identified the restriction map $H^0(X, L^n) \otimes_{\mathbb{C}} \mathcal{O}_{X_0} \to L_0^n$, you will have then completed this step.