Bin packing when items can be broken
Here is a proof that the problem remains hard when $k=1$. For simplicity it uses an item of size equal to the bin capacity. If you require each item to be smaller than the bin capacity you can subtract a small enough constant from this item's size.
Consider an instance of $3$-partition in which $3n$ positive integers  $x_1, x_2, \dots, x_{3n}$ of total sum $S$ need to be partitioned into $n$ triples $t_1, \dots, t_n$, each with total sum at most $T=\frac{S}{n}$, where $T$ is also an integer.
Assume that $x_i \ge \frac{2T}{7}$ (it is known that the problem remains strongly NP-complete in this case).
Furthermore assume that $n &gt; \max\{8,T+1\}$ (if this is not the case, multiply each integer in the instance by $10$, add many copies of the integers in $\{10T-2, 1, 1\}$ to the input, and then convert the resulting instance into an instance that satisfies $x_i \ge \frac{2T}{7}$. This conversion multiplies the new value of $T$ by $7$).
Consider the instance of your version of bin-packing in which $k=1$ and you need to pack the set of elements $Y = \{y_1, \dots, y_m\}$ into the fewest number of bins of capacity $n T$ (you can set the capacity of each bin to $1$ by a suitable scaling of all the involved quantities).
Let $m=3n+1$, $y_m = nT$ and $y_i = (n-1) x_i$ for $i=1,\dots,3n$.
If the $3$-partition instance is a yes-instance then $Y$ can be packed into $n$ bins.
Create a bin $b = (y_i, y_j, y_h)$ for each triple $t = (x_i, x_j, x_h)$ of a solution of $3$-partition. Notice that $y_i + y_j + y_h = (n-1) (x_i + x_j +x_h) = (n-1)T$, leaving $b$ with unused capacity of $T$.
The only element not assigned to a bin is $y_m$. Split $y_m = nT$ into $n$ pieces of size $T$ each, and add a piece to each of the $n$ bins.
If $Y$ can be packed into $n$ bins then the $3$-partition instance is a yes-instance.
Notice that $\sum_{i=1}^m y_i = (n-1) \sum_{i=1}^{3n}x_i + nT = (n-1)nT + nt = n \cdot nT$, showing that each each of the $n$ bins must be completely full.
We can assume w.l.o.g., that if some element is split into pieces, this element is $y_m$.
Indeed, if an element $y' \neq y_m$ is split, there must be one bin $b'$ that is used entirely for $y_m$. We can then swap each piece of $y'$ with a piece of $y_m$, so that $y'$ is now entirely contained in $b'$.
Notice now that each bin $b$ must contain exactly $3$ elements from $\{ y_1, \dots, y_{3n}\}$.
Indeed, if this was not true there would exist a bin that contains at least $4$ elements from $\{ y_1, \dots, y_{3n}\}$ and its used capacity would be at least
$$
4(n-1)\left(\frac{2T}{7}\right) = \left( \frac{8}{7}n - \frac{8}{7} \right)T &gt; nT.
$$
We construct a solution for the $3$-partition instance by creating a triple $t = (x_i, x_j, x_h)$ for each bin $b$ containing $y_i, y_j, y_h$ with $i,j,h \le 3n$.
The total weight of $t$ is at most:
$$
x_i + x_j + x_h = \frac{y_i + y_j + y_h}{n-1} \le \frac{nT}{n-1} 
= \frac{(n-1)T+T}{n-1} = T + \frac{T}{n-1} &lt; T+1,
$$
i.e., $x_i + x_j + x_h \le T$.