Let $f: X\longrightarrow \operatorname{Spec} A$ be a morphism and $\mathcal{F}$ be a coherent sheaf on $X$, flat over $\operatorname{Spec} A$. If $M$ is an $A$-algebra then, as in Hartshrone III $\S 12$ there is a natural mapping $$H^0(X, \mathcal{F})\otimes M \longrightarrow H^0(X, \mathcal{F}\otimes_A M).$$ I am trying to understand for any $x\in H^0(X, \mathcal{F})$ what is the image of $x\otimes 1$. My guess is the following.
Since the sheaf $\mathcal{F}\otimes_A M$ is exactly the sheaf $\mathcal{F}\otimes_{\mathcal{O}_X} f^*\widetilde{M}$, so we can define the above mapping \begin{align} H^0(X, \mathcal{F})\otimes M &\longrightarrow H^0(X, \mathcal{F}\otimes_A M) \ as\\ x\otimes 1 & \longrightarrow (x_P\otimes 1)_{P\in X}. \end{align}
My reason is the following, since we know that for any presheaf $\mathcal{F}$ on $X$, the global section of the sheafification $\mathcal{F}^+$ is given by $$\mathcal{F}^+(X)= \{(x_P\in \mathcal{F}_P)_{P\in X}; \ \mathrm{with\ some\ compatible\ condition}\}.$$ So, we can define the mapping as above.