I am not sure what kind of interpretation you want, but heuristically, $\mathcal{F}_{t\wedge\tau_H}$ is the information we have from observing the Brownian motion up until the time $t\wedge\tau_H$. Then $E[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]=P(t\le\tau_K\mid\mathcal{F}_{t\wedge\tau_H})$ is the probability that $B_t$ is still in $K$, given this information.

As for $E[1_EE[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]]$, one thing you might observe (and again this may not be the kind of thing you are looking for) is the following:
\begin{align*}
E[1_EE[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]]
&= E[E[1_EE[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]
\mid\mathcal{F}_{t\wedge\tau_H}]]\\\\
&= E[E[1_E\mid\mathcal{F}_{t\wedge\tau_H}]
E[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]]\\\\
&= E[P(E\mid\mathcal{F}_{t\wedge\tau_H})
P(t\le\tau_K\mid\mathcal{F}_{t\wedge\tau_H})].
\end{align*}
So informally, you observe the Brownian path until time $t\wedge\tau_H$, compute the conditional probabilities of $E$ and $\{t\le\tau_K\}$, multiply them together, and then average the result over all such observations.

**Edit:**

Perhaps I have misunderstood your question. I wanted to address your comment, "By $E[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]$ I mean the $\mathcal{F}_{t\wedge\tau_H}$-measurable r.v. It is not a probability anymore." I am not yet permitted by the software to post comments of my own, so I must address it as an edit to my answer.

The r.v. $E[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}]$ is a [0,1]-valued, measurable function of the Brownian path up until time $t\wedge\tau_H$. Given such a path, say $\omega$, the number
$E[1_{\{t\le\tau_K\}}\mid\mathcal{F}_{t\wedge\tau_H}] (\omega)\in[0,1]$ is generally interpreted (albeit informally) as the probability that $t\le\tau_K$, given that the Brownian path was $\omega$. This was the genesis of the final comment in my original answer.