If $\mathcal{X}$ and $\mathcal{Y}$ are representable by algebraic spaces, then this is also equivalent to $f$ (as a morphism of algebraic spaces) having property $\mathcal{P}$. If $\mathcal{P}$ is also preserved under any base change, and fppf local on the base, then for morphisms $f$ which are representable by algebraic spaces this is also equivalent to $f$ having property $\mathcal{P}$ in the sense of Properties of Stacks, Section 100.3.
Proof.
Let us prove the implication (1) $\Rightarrow $ (2). Pick an algebraic space $V$ and a surjective and smooth morphism $V \to \mathcal{Y}$. Pick an algebraic space $U$ and a surjective and smooth morphism $U \to \mathcal{X} \times _\mathcal {Y} V$. Note that $U \to \mathcal{X}$ is surjective and smooth as well, as a composition of the base change $\mathcal{X} \times _\mathcal {Y} V \to \mathcal{X}$ and the chosen map $U \to \mathcal{X} \times _\mathcal {Y} V$. Hence we obtain a diagram as in (1). Thus if (1) holds, then $h : U \to V$ has property $\mathcal{P}$, which means that (2) holds as $U \to \mathcal{X}$ is surjective.
Conversely, assume (2) holds and let $U, V, a, b, h$ be as in (2). Next, let $U', V', a', b', h'$ be any diagram as in (1). Picture
\[ \xymatrix{ U \ar[d] \ar[r]_ h & V \ar[d] \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} } \quad \quad \xymatrix{ U' \ar[d] \ar[r]_{h'} & V' \ar[d] \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} } \]
To show that (2) implies (1) we have to prove that $h'$ has $\mathcal{P}$. To do this consider the commutative diagram
\[ \xymatrix{ U \ar[dd]^ h & U \times _\mathcal {X} U' \ar[d] \ar[l] \ar@/^6ex/[dd]^{(h, h')} \ar[r] & U' \ar[dd]^{h'} \\ & U \times _\mathcal {Y} V' \ar[lu] \ar[d] & \\ V & V \times _\mathcal {Y} V' \ar[l] \ar[r] & V' } \]
of algebraic spaces. Note that the horizontal arrows are smooth as base changes of the smooth morphisms $V \to \mathcal{Y}$, $V' \to \mathcal{Y}$, $U \to \mathcal{X}$, and $U' \to \mathcal{X}$. Note that
\[ \xymatrix{ U \times _\mathcal {X} U' \ar[d] \ar[r] & U' \ar[d] \\ U \times _\mathcal {Y} V' \ar[r] & \mathcal{X} \times _\mathcal {Y} V' } \]
is cartesian, hence the left vertical arrow is smooth as $U', V', a', b', h'$ is as in (1). Since $\mathcal{P}$ is smooth local on the target by Descent on Spaces, Lemma 74.20.2 part (2) we see that the base change $U \times _\mathcal {Y} V' \to V \times _\mathcal {Y} V'$ has $\mathcal{P}$. Since $\mathcal{P}$ is smooth local on the source by Descent on Spaces, Lemma 74.20.2 part (1) we can precompose by the smooth morphism $U \times _\mathcal {X} U' \to U \times _\mathcal {Y} V'$ and conclude $(h, h')$ has $\mathcal{P}$. Since $V \times _\mathcal {Y} V' \to V'$ is smooth we conclude $U \times _\mathcal {X} U' \to V'$ has $\mathcal{P}$ by Descent on Spaces, Lemma 74.20.2 part (3). Finally, since $U \times _ X U' \to U'$ is surjective and smooth and $\mathcal{P}$ is smooth local on the source (same lemma) we conclude that $h'$ has $\mathcal{P}$. This finishes the proof of the equivalence of (1) and (2).
If $\mathcal{X}$ and $\mathcal{Y}$ are representable, then Descent on Spaces, Lemma 74.20.3 applies which shows that (1) and (2) are equivalent to $f$ having $\mathcal{P}$.
Finally, suppose $f$ is representable, and $U, V, a, b, h$ are as in part (2) of the lemma, and that $\mathcal{P}$ is preserved under arbitrary base change. We have to show that for any scheme $Z$ and morphism $Z \to \mathcal{X}$ the base change $Z \times _\mathcal {Y} \mathcal{X} \to Z$ has property $\mathcal{P}$. Consider the diagram
\[ \xymatrix{ Z \times _\mathcal {Y} U \ar[d] \ar[r] & Z \times _\mathcal {Y} V \ar[d] \\ Z \times _\mathcal {Y} \mathcal{X} \ar[r] & Z } \]
Note that the top horizontal arrow is a base change of $h$ and hence has property $\mathcal{P}$. The left vertical arrow is smooth and surjective and the right vertical arrow is smooth. Thus Descent on Spaces, Lemma 74.20.3 kicks in and shows that $Z \times _\mathcal {Y} \mathcal{X} \to Z$ has property $\mathcal{P}$.
$\square$
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