= Solution
Take the <projective line> $X=\mathbb P^1_k$ and the <coherent sheaves> $\mathcal F=\mathcal O_X$ and $\mathcal G=\mathcal O_X(-1)$, where the latter is a <twisting sheaf on projective space>. Both are <line bundles>, so each <stalk> is a free rank-one $\mathcal O_{X,x}$-<module> and therefore $\mathcal F_x\simeq\mathcal G_x$ at every point.
They are nevertheless not isomorphic. On the standard charts with coordinates $t$ and $t^{-1}$, choose frames $e_0,e_1$ of $\mathcal O(-1)$ with $e_1=t^{-1}e_0$. A <global section> would give polynomials $a(t)$ and $b(t^{-1})$ satisfying
$$
a(t)=t^{-1}b(t^{-1}).
$$
The left side has only nonnegative powers of $t$, and the right side only strictly negative powers. Thus both vanish, giving $\Gamma(X,\mathcal O(-1))=0$. For the <structure sheaf>, the corresponding equality is $a(t)=b(t^{-1})$, whose solutions are precisely the constants. Hence
$$
\boxed{H^0(X,\mathcal F)=k,\qquad H^0(X,\mathcal G)=0,\qquad\mathcal F\not\simeq\mathcal G.}
$$
A <morphism of sheaves> which is an isomorphism on all <stalks> would be an isomorphism; this example only asserts the existence of separate <stalk> isomorphisms, not that they come from one <morphism of sheaves>.
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