$$This post answers a question that had come up some time ago with Arthur Apter, and more recently with Philipp Schlicht and Arthur Apter.

Definition. A cardinal $\kappa$ is uniformly $<\theta$-supercompact if there is an embedding $j:V\to M$ having critical point $\kappa$, with $j(\kappa )>\theta$ and $M^{<\theta }\subset M$.

(Note:  This is typically stronger than merely asserting that $\kappa$ is $\gamma$-supercompact for every $\gamma <\theta$, a property which is commonly denoted $<\theta$-supercompact, so I use the adjective “uniformly” to highlight the distinction.)

Two easy observations are in order.  First, if $\theta$ is singular, then $\kappa$ is uniformly $<\theta$-supercompact if and only if $\kappa$ is $\theta$-supercompact, since the embedding $j:V\to M$ will have $j{}^{\prime \prime }\lambda \in M$ for every $\lambda <\theta$, and we may assemble $j{}^{\prime \prime }\theta$ from this inside $M$, using a sequence of length $\text{cof}(\theta )$. Second, in the successor case, $\kappa$ is uniformly $<{\lambda }^{+}$-supercompact if and only if $\kappa$ is $\lambda$-supercompact, since if $j:V\to M$ has $M^{\lambda }\subset M$, then it also has $M^{<{\lambda }^{+}}\subset M$. So we are mainly interested in the concept of uniform $<\theta$-supercompactness when $\theta$ is weakly inaccessible.

Definition. Let us say of a cardinal $\kappa$ that $⟨{\mu }_{\lambda }\mid \lambda <\theta ⟩$ is a coherent $\theta$-system of normal fine measures, if each ${\mu }_{\lambda }$ is a normal fine measure on $P_{\kappa }\lambda$, which cohere in the sense that if $\lambda <\delta <\theta$, then ${\mu }_{\lambda }{\le }_{RK}{\mu }_{\delta }$, and more specifically $X\in {\mu }_{\lambda }$ if and only if $\{\sigma \in P_{\kappa }\delta \mid \sigma \cap \lambda \in X\}\in {\mu }_{\delta }$.  In other words, ${\mu }_{\lambda }=f\ast {\mu }_{\delta }$, where $f:P_{\kappa }\delta \to P_{\kappa }\lambda$ is the function that chops off at $\lambda$, so that $f:\sigma \mapsto \sigma \cap \lambda$.

Theorem.  The following are equivalent, for any regular cardinals $\kappa \le \theta$.

1. The cardinal $\kappa$ is uniformly $<\theta$-supercompact.

2. There is a coherent $\theta$-system of normal fine measures for $\kappa$.

Proof. The forward implication is easy, since if $j:V\to M$ has $M^{<\theta }\subset M$, then we may let ${\mu }_{\lambda }$ be the normal fine measure on $P_{\kappa }\lambda$ generated by $j{}^{\prime \prime }\lambda$ as a seed, so that $X\in {\mu }_{\lambda }⟺j{}^{\prime \prime }\lambda \in j(X)$.  Since the seeds $j{}^{\prime \prime }\lambda$ cohere as initial segments, it follows that ${\mu }_{\lambda }{\le }_{RK}{\mu }_{\delta }$ in the desired manner whenever $\lambda <\delta <\theta$.

Conversely, fix a coherent system $⟨{\mu }_{\lambda }\mid \lambda <\theta ⟩$ of normal fine measures. Let $j_{\lambda }:V\to M_{\lambda }$ be the ultrapower by ${\mu }_{\lambda }$. Every element of $M_{\lambda }$ has the form $j_{\lambda }(f)(j{}^{\prime \prime }\lambda )$.  Because of coherence, we have an elementary embedding $k_{\lambda ,\delta }:M_{\lambda }\to M_{\delta }$ defined by $$k_{\lambda ,\delta }:j_{\lambda }(f)(j{}^{\prime \prime }\lambda )\mapsto j_{\delta }(f)(j{}^{\prime \prime }\lambda ).$$ It is not difficult to check that these embeddings altogether form a commutative diagram, and so we may let $j:V\to M$ be the direct limit of the system, with corresponding embeddings $k_{\lambda ,\theta }:M_{\lambda }\to M$.  The critical point of $k_{\lambda ,\delta }$ and hence also $k_{\lambda ,\theta }$ is larger than $\lambda$.  This embedding has critical point $\kappa$, and I claim that $M^{<\theta }\subset M$. To see this, suppose that $z_{\alpha }\in M$ for each $\alpha <\beta$ where $\beta <\theta$.  So $z_{\alpha }=k_{{\lambda }_{\alpha },\theta }(z_{\alpha }^{\ast })$ for some $z_{\alpha }^{\ast }\in M_{{\lambda }_{\alpha }}$. Since $\theta$ is regular, we may find $\lambda <\theta$ with ${\lambda }_{\alpha }\le \lambda$ for all $\alpha <\beta$ and also $\beta \le \lambda$, and so without loss we may assume ${\lambda }_{\alpha }=\lambda$ for all $\alpha <\beta$. Since $M_{\lambda }$ is closed under $\lambda$-sequences, it follows that ${\overrightarrow{z}}^{\ast }=⟨z_{\alpha }^{\ast }\mid \alpha <\beta ⟩\in M_{\lambda }$.  Applying $k_{\lambda ,\theta }$ to ${\overrightarrow{z}}^{\ast }$ gives precisely the desired sequence $\overrightarrow{z}=⟨z_{\alpha }\mid \alpha <\beta ⟩$ inside $M$, showing this instance of $M^{<\theta }\subset M$. QED

The theorem does not extend to singular $\theta$.

Theorem.  If $\kappa$ is $\theta$-supercompact for a singular strong limit cardinal $\theta$ above $\kappa$, then there is a transitive inner model in which $\kappa$ has a coherent system $⟨{\mu }_{\lambda }\mid \lambda <\theta ⟩$  of normal fine measures, but $\kappa$ is not uniformly $<\theta$-supercompact.

Thus, the equivalence of the first theorem does not hold generally for singular $\theta$.

Proof.  Suppose that $\kappa$ is $\theta$-supercompact, where $\theta$ is a singular strong limit cardinal. Let $j:V\to M$ be a witnessing embedding, for which $\kappa$ is not $\theta$-supercompact in $M$ (use a Mitchell-minimal measure).  Since $\theta$ is singular, this means by the observation after the definition above that $\kappa$ is not uniformly $<\theta$-supercompact in $M$. But meanwhile, $\kappa$ does have a coherent system of normal fine ultrafilters in $M$, since the measures defined by $X\in {\mu }_{\lambda }⟺j{}^{\prime \prime }\lambda \in j(X)$ form a coherent system just as in the theorem, and the sequence $⟨{\mu }_{\lambda }\mid \lambda <\theta ⟩$ is in $M$ by $\theta$-closure. QED

The point is that in the singular case, the argument shows only that the direct limit is $<\text{cof}(\theta )$-closed, which is not the same as $<\theta$-closed when $\theta$ is singular.

The example of singular $\theta$ also shows that $\kappa$ can be $<\theta$-supercompact without being uniformly $<\theta$-supercompact, since the latter would imply full $\theta$-supercompactness, when $\theta$ is singular, but the former does not. The same kind of reasoning separates uniform from non-uniform $<\theta$-supercompactness, even when $\theta$ is regular.

Theorem. If $\kappa$ is uniformly $<\theta$-supercompact for an inaccessible cardinal $\theta$, then there is a transitive inner model in which $\kappa$ is $<\theta$-supercompact, but not uniformly $<\theta$-supercompact.

Proof. Suppose that $\kappa$ is uniformly $<\theta$-supercompact, witnessed by embedding $j:V\to M$, with $M^{<\theta }\subset M$, and furthermore assume that $j(\kappa )$ is as small as possible among all such embeddings. It follows that there can be no coherent $\theta$-system of normal fine measures for $\kappa$ inside $M$, for if there were, the direct limit of the associated embedding would send $\kappa$ below $j(\kappa )$, which from the perspective of $M$ is a measurable cardinal far above $\kappa$ and $\theta$. But meanwhile, $\kappa$ is $\beta$-supercompact in $M$ for every $\beta <\theta$. Thus, $\kappa$ is $<\theta$-supercompact in $M$, but not uniformly $<\theta$-supercompact, and so the notions do not coincide. QED

Meanwhile, if $\theta$ is weakly compact, then the two notions do coincide. That is, if $\kappa$ is $<\theta$-supercompact (not necessarily uniformly), and $\theta$ is weakly compact, then in fact $\kappa$ is uniformly $<\theta$-supercompact, since one may consider a model $M$ of size $\theta$ with $\theta \in M$ and $V_{\theta }\subset M$, and apply a weak compactness embedding $j:M\to N$. The point is that in $N$, we get that $\kappa$ is actually $\theta$-supercompact in $N$, which provides a uniform sequence of measures below $\theta$.