Step $-1$ (unnecessary). Show that the set of sections Hom$(\mathscr F|_U,\mathscr G|_U)$ over an open subset $U\subset X$ is an abelian group (so that $\mathcal Hom(\mathscr F,\mathscr G)$ will be a sheaf of abelian groups). This is easy.
Step $0$. $U\mapsto \textrm{Hom}(\mathscr F|_U,\mathscr G|_U)$ is a presheaf (note: as pointed out in a comment, a section of this presheaf is a morphism of sheaves!). The restriction is defined as follows: for fixed $U$, and an open subset $V\subset U$, a section $\sigma\in\textrm{Hom}(\mathscr F|_U,\mathscr G|_U)$ goes to $\sigma|_V\in \textrm{Hom}(\mathscr F|_V,\mathscr G|_V)$, where $\sigma|_V$ is the morphism of sheaves on $V$ defined by $\sigma|_V(W)=\sigma(W):\mathscr F(W)\to\mathscr G(W)$ for any open subset $W\subset V$ (which is also open in $U$! for this reason, the squares that must commute, over $V$, do commute because they already commuted over $U$).
Step $1$. The first sheaf axiom. Let $U=\bigcup_{i\in I} U_i$ be an open covering of an open subset $U\subset X$. Let $\sigma: \mathscr F|_U\to\mathscr G|_U$ be a section such that $\sigma_i:=\sigma|_{U_i}=0$ for all $i\in I$. We want to show that $\sigma=0$.
Let $g\in\mathscr F(U)$ be a fixed section. Then look at the (zero!) morphisms of abelian groups
$$
\sigma_i(U_i):\mathscr F(U_i)\to\mathscr G(U_i) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,g_i\mapsto 0.
$$
Now, because $\mathscr{G}$ is a sheaf and the image of the section $g$ under $\sigma(U)$ restricts to zero on every open set in an open cover $\{U_i \}$ (recall the usual commutative diagram for morphisms of (pre)sheaves/natural transformations), it has the property
$$
\sigma(U)(g)=0.
$$
Because this holds for every $g\in\mathscr F(U)$, we conclude that $\sigma(U)=0$, hence $\sigma=0$, as claimed.
Step 2. The second sheaf axiom.
Let again $U=\bigcup_{i\in I} U_i$ be an open covering of an open subset $U\subset X$, and let $\{\phi_i:\mathscr F|_{U_i}\to\mathscr G|_{U_i}\}_{i\in I}$ be a family of sections such that $\phi_i=\phi_j$ on $U_{ij}$. We want a global $\phi$ (section over $U$) such that $\phi|_{U_i}=\phi_i$.
If $V\subset U$, then $A_i:=U_i\cap V$ cover $V$. So let us fix a section $g\in \mathscr F(V)$ and let us set $g_i:=g|_{A_i}$. We can give a name (say $t_i$) to the image of $g_i$ under $\phi(A_i)$, namely
$$
\phi_i(A_i):\mathscr F(A_i)\to\mathscr G(A_i) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,g_i\mapsto t_i.
$$
The compatibility of the $\phi_i$'s implies that of the $t_i$'s, and since $\mathscr G$ is a sheaf there exists a global section $t\in \mathscr G(V)$ such that $t|_{A_i}=t_i$ for every $i$. We can define the $\phi$ that we are looking for by
$$
\phi(V):\mathscr F(V)\to\mathscr G(V) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, g\mapsto t.
$$
for every $V\subset U$. In this way, by construction, $\phi|_{U_i}=\phi_i$, as wanted.
If $(X,\mathcal O_X)$ is a scheme and if $U\subset X$ is is an open affine subscheme, then we have for all quasi-coherent sheaves $\mathcal F,\mathcal G$ of $\mathcal O_X$-modules the extremely pleasant equality of $\mathcal O_X(U)$-modules:
$$(\mathcal F \otimes_{ \mathcal O_X} \mathcal G)(U)= \mathcal F(U) \otimes_{\mathcal O_X(U)} \mathcal G(U) $$
However if $U$ is not affine all hell can break loose:
For example if $X=\mathbb P^1_\mathbb C$ is the complex projective line and if $\mathcal F=\mathcal O_X(1), \mathcal G=\mathcal O_X(-1)$, then for these quasi-coherent sheaves $\mathcal F \otimes_{ \mathcal O_X} \mathcal G=\mathcal O_X$, so that for $U=X$:
$$(\mathcal F \otimes_{ \mathcal O_X} \mathcal G)(X)= \mathcal O_X(X) =\mathbb C \neq \mathcal F(X)\otimes _{\mathcal O_X(X)}\mathcal G(X)=\mathbb C^2\otimes_\mathbb C 0=0$$
Reference
The first displayed equality is proved in Qing-Liu, chapter 5, Proposition 1.12 (b), page 162.
Best Answer
Hom is the global sections of sheaf Hom (and so it contains strictly less information). Part of the difference between derived Hom and derived sheaf Hom has to do with derived global sections.