Let $\{H_i: i \in I\}$ be a collection of Hilbert spaces. We can form the Hilbert space direct sum $$H:= \bigoplus_{i \in I} H_i$$
Question: Is there a "nice" dense subset of $B_0(H)$ (compact operators on the direct sum $H$)? For example, something like the operators $$\left\{\bigoplus_{i \in I} T_i \ \Bigg| \ T_i \in F(H_i), \text{all but finitely many $T_i$ are zero}\right\}$$
where $F(H_i)$ are the finite-rank operators on $H_i$.
Let me begin with a discussion of "operators vs. matrices", striving to make precise statements based on adjoints and compositions of operators, while avoiding the need to make identifications between sets, such as between a factor of a direct sum and its canonical image.
Let $\{H_i\}_{ i \in I}$ be a collection of Hilbert spaces and put $$ H:= \bigoplus_{i \in I} H_i. $$ For each $i$, denote by $\iota _i:H_i\to H$ the inclusion map and observe that $\iota _i^*$ is precisely the projection from $H$ onto $H_i$. Incidentally, since $\iota _i$ is an isometry, then $\iota _i^*\iota _i$ is the identity on the domain of $\iota _i$ , while $\iota _i\iota _i^*$ is the orthogonal projection onto $\iota_i (H_i)$, a.k.a. the canonical image of $H_i$ in $H$. See $(\dagger)$ below regarding a subtle conceptual distinction between $\iota _i\iota _i^*$ and $\iota _i^*$.
Given any bounded linear operator $T\in B(H)$, put $$ T_{i,j}=\iota _i^*T\iota _j, \quad \forall i,j\in I, $$ so that each $T_{i,j}\in B(H_j,H_i)$. The "matrix" $(T_{i,j})_{i,j}$ may then be seen as the matrix of $T$ relative to the above decomposition of $H$, and it can be used to recover $T$ in the following way: given any vector $\xi \in H$, write $\xi =(\xi _i)_{i\in I}$, so that each $\xi _i=\iota _i^*\xi $. Noticing that $\xi =\sum_{j\in I} \iota _j\xi _j$, one then has for every $i$ that $$ \iota _i^*T\xi = \sum_{j\in I} \iota _i^*T\iota _j\xi _j, $$ so, writing $T(\xi )=(\eta _i)_{i\in I}$, we see that the $i^{th}$ component $\eta _i$ of $T(\xi )$ may be computed in the familiar way as $$ \eta _i=\sum_{j\in I} T_{i, j}\xi _j. $$
Conversely, if $(T_{i,j})_{i,j}$ is any given matrix, with $T_{i,j}\in B(H_j,H_i)$, assume that $(T_{i,j})_{i,j}$ has finitely many nonzero entries. One can then define a bounded operator $T$ on $H$ by $$ T = \sum_{i, j} \iota _iT_{i,j}\iota^* _j, \tag 1 $$ and it is easy to see that the matrix of $T$ coincides with the originally given matrix. In case one moreover has that each $T_{i,j}$ lies in the set $K(H_j,H_i)$ of compact operators, then $T$ is clearly compact. If instead each $T_{i,j}$ lies in the set $F(H_j,H_i)$ of finite rank operators, then $T$ has finite rank.
Directly addressing the question posed by the OP, the set formed by all operators built as in (1), from finitely supported matrices $(T_{i,j})_{i,j}$, with each $T_{i,j}\in F(H_j,H_i)$, is a dense subspace of $K(H)$. The proof may be easily done by observing that the identity operator $1_H$ may be writen as $$ 1_H = \sum_i \iota _i\iota _i^*, $$ with convergence in the strong operator topology, and moreover noticing that left- or right-multiplication by compact operators turns strong convergence into norm convergence.
$(\dagger)$ There is a subtle conceptual distinction between $\iota _i\iota _i^*$ and $\iota _i^*$ in the sense that both project onto $H_i$, but the former has $H$ for co-domain while the latter's co-domain is $H_i$. Co-domains are often irrelevant in Math, but since the co-domain of an operator becomes the domain of its adjoint, it is sometimes important to pay attention to co-domains too! In this note all compositions $TS$ of operators are such that the domain of $T$ is equal to the co-domain of $S$, even though for $TS$ to make sense, it is enough for the domain of $T$ to contain the range of $S$.