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user36938
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Here is a coordinate-free version. Let $(V, \psi)$ be a 4-dimensional symplectic space over a field, and let $\omega \in (\wedge^2 V)^{\ast} = \wedge^2(V^{\ast})$ be the nonzero 2-form arising from $\psi$. Then on the 6-dimensional vector space $W = \wedge^2(V)$ the kernel of $\omega$ is a hyperplane $H$ of dimension 5, and on $W$ there is a natural non-degenerate quadratic form $q$ valued in the line $\wedge^4(V)$ via $q(w) = (1/2)(w \wedge w)$ for $w \in W$ (e.g., if $w = e_1 \wedge e_2 + e_3 \wedge e_4$ for a basis $\{e_i\}$ of $W$ then $q(w) = e_1 \wedge e_2 \wedge e_3 \wedge e_4$); the definition of $q$ uses base change from the $\mathbf{Z}_{(2)}$-flat case for $(V,\psi)$ if $2$ isn't a unit.

By computing in linear coordinates of $V$ that "standardize" $\psi$ we see that $q$ is a split quadratic form on $W$, and the action of ${\rm{SL}}(V)$ on $W$ clearly preserves $q$ while the action of its subgroup ${\rm{Sp}}(V,\psi)$ preserves $H$. Hence, the action on $H$ defines a map $${\rm{Sp}}(V,\psi) \rightarrow {\rm{O}}(q|_H),$$ so this lands inside ${\rm{SO}}(q|_H)$ and as such defines a homomorphism $${\rm{Sp}}_4 = {\rm{Sp}}(V,\psi) \rightarrow {\rm{SO}}(q|_H) = {\rm{SO}}_5.$$

This map kills the center $\mu_2$ inside ${\rm{Sp}}_4$, and thereby identifies ${\rm{Sp}}_4$ as the degree-2 "simply connected" central cover of ${\rm{SO}}_5$ (in the sense of algebraic groups). Hence, this uniquely lifts to an isomorphism of ${\rm{Sp}}_4$ onto ${\rm{Spin}}_5$.

A nice feature of this conceptual construction is that it kills two birds with one stone: if we don't restrict to $H$ and instead work with the entire 6-dimensional $W$ then a similar construction defines the isomorphism of ${\rm{SL}}_4 = {\rm{SL}}(V)$ onto ${\rm{Spin}}_6$.

user36938
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