**UPDATE.** My original answer addressed the case when $t(L)$ is defined as the minimal binary subtree (which may be rooted, i.e. have one vertex of degree 2). This is now posed as Case 1.

Case 2 down below addresses the case when $t(L)$ is the minimal arbitrary subtree (i.e., connected acyclic graph).

**Case 1. $t(L)$ is a binary subtree.**

Each valid partition corresponds to a binary subtree of the given tree. Namely, let $T'$ be a binary subtree of $T$. If $T'$ consists of a single edge $t$, then this edge defines of a $2$-partition of $L(T)$ obtained by removal of $t$ from $T$, which splits $L(T)$ into two complementary subsets.

More generally, $T'$ with $k>2$ leaves defines a $k$-partition of $L(T)$ as follows. Every leaf $t$ of $T'$ corresponds to a subset of $L(T)$: removal of the leaf edge of $t$ in $T'$ from $T$ splits $T$ into two subgraphs, one of which is edge-disjoint with $T'$. The set of leaves of this subgraph defines a subset $L_t$ of $L(T)$.
It is clear that $L_{t_1}$ and $L_{t_2}$ are disjoint for any distinct leaves $t_1$ and $t_2$ of $T'$. Furthermore, each leaf of $T$ belongs to $L_t$ for some leaf $t$ of $T$. That is, the sets $L_t$, where $t$ runs over the leaves of $T'$, form a valid $k$-partition of $L(T)$.

Here is a modification of the algorithm from Counting the number of subgraphs in a given labeled tree that counts the number of binary subtrees with $k$ leaves in the given binary $T$.

Let $\ell$ be a fixed leaf of $T$ and $r$ be its the only neighbor. Let $T_r$ be a rooted binary tree obtained from $T$ by removal of $\ell$, with the root at $r$. More generally, let $T_v$ denote the subtree of $T_r$, rooted at vertex $v$.

For a vertex $v$ in $T_r$, define
$$A_v(z)=a_1\cdot z+a_2\cdot z^2+\dots,$$
$$B_v(z)=b_1\cdot z+b_2\cdot z^2+\dots,$$
where $a_i$ is the number of binary subtrees of $T_v$ that have $i$ leaves and include $v$; and similarly, $b_i$ is the number of binary subtrees of $T_v$ that have $i$ leaves and do not include $v$.

If $v$ is a leaf then $A_v(z)=z$ and $B_v(z)=0$.

If $u_1,u_2$ are the children of a non-leaf $v$, then
$$
\begin{cases}
A_v(z) = z + A_{u_1}(z)\cdot A_{u_2}(z), \\
B_v(z) = A_{u_1}(z) + B_{u_1}(z) + A_{u_2}(z) + B_{u_2}(z),
\end{cases}
$$
and the answer is the coefficient of $z^{k-1}$ in
$$A_r(z)+B_r(z).$$
Here we account for the fact that rooted binary subtrees in $T_r$ with $k-1$ leaves correspond to unrooted binary subtrees in $T$ with $k$ leaves, namely, a subtree with a root $r'$ in $T_r$ gets the "parent" edge of $r'$ attached in $T$ (in particular, for $r'=r$, we get an extra leaf $\ell$ in the corresponding subtree of $T$).

The two recurrences need to be applied once for each vertex of $T_r$ in the bottom-up fashion, starting from the leaves of $T_r$ and ending at the root $r$.

P.S. Elements of valid partitions of $L(T)$ are studied to some extent in my paper under the name of *$T$-consistent multicolors*.

**Case 2. $t(L)$ is an arbitrary subtree.**

Smilarly to the above, we define $\ell$, $r$, and $T_v$.

For a vertex $v$ in $T_r$, define
$$A_v(z)=a_1\cdot z+a_2\cdot z^2+\dots,$$
$$B_v(z)=b_1\cdot z+b_2\cdot z^2+\dots,$$
$$C_v(z)=c_1\cdot z+c_2\cdot z^2+\dots,$$
where

$a_i$ is the number of forests in $T_v$, containing $v$ and consisting of $t(L_j)$, where $L_j$ ($j=1,2,\dots,i$) form an $i$-partition of $L(T_v)$ (notice that a subtree containing $v$ must also contain both its children in $T_v$ if there are any);

$b_i$ is the number of forests in $T_v$, not containing $v$ and consisting of $t(L_j)$, where $L_j$ ($j=1,2,\dots,i$) form a partition of $L(T_v)$;

$c_i$ is the number of forests in $T_v$, containing $v$ as a leaf and consisting of $t(L_1),\dots, t(L_{i-1})$ and $t(L_i\cup\{v\})$, where $L_j$ ($j=1,2,\dots,i$) form an $i$-partition of $L(T_v)$ (notice that a subtree containing $v$ must also contain exactly one child of $v$ in $T_v$).

If $v$ is a leaf then $A_v(z)=z$ and $B_v(z)=C_v(z)=0$.

If $u_1,u_2$ are the children of a non-leaf $v$, then
$$
\begin{cases}
A_v(z) = \frac{1}{z}\cdot (A_{u_1}(z)+C_{u_1}(z))\cdot (A_{u_2}(z)+C_{u_2}(z)), \\
B_v(z) = (A_{u_1}(z)+B_{u_1}(z))\cdot (A_{u_2}(z)+B_{u_2}(z)), \\
C_v(z) = (A_{u_1}(z)+C_{u_1}(z))\cdot (A_{u_2}(z)+B_{u_2}(z))+(A_{u_1}(z)+B_{u_1}(z))\cdot (A_{u_2}(z)+C_{u_2}(z)).
\end{cases}
$$

The number of valid $k$-partitions of $T$ in given by the coefficient of $z^k$ in
$$(1+z)\cdot A_r(z) + z\cdot B_r(z) + C_r(z).$$

*Example.* Let $T$ be a tree on 6 vertices $a,b,c,d,e,f$ with edges $(a,b)$, $(b,c)$, $(c,d)$, $(b,e)$, $(c,f)$. So the leaves of $T$ are $a,d,e,f$.

First we fix a leaf $\ell$, say, $\ell = a$. Then $r=b$ and we have a rooted tree $T_r$ with leaves $d,e,f$. For each of them we have
$$(A_d(z),B_d(z),C_d(z)) = (A_e(z),B_e(z),C_e(z)) = (A_f(z),B_f(z),C_f(z)) = (z,0,0).$$

From ABC-values at $d,f$, we compute them at their parent $c$:
$$(A_c(z),B_c(z),C_c(z)) = (z,z^2,2z^2).$$

From ABC-values at $c,e$, we compute them at their parent $r=b$:
$$(A_r(z),B_r(z),C_r(z)) = (z+2z^2,z^2+z^3,2z^2+3z^3).$$

Now we compute the answer:
$$(1+z)\cdot A_r(z) + z\cdot B_r(z) + C_r(z) = (1+z)\cdot (z+2z^2)+z\cdot (z^2+z^3)+(2z^2+3z^3)$$
$$=z+5z^2+6z^3+z^4.$$
It is easy to check that we indeed have one valid 1-parition of $L(T)$, five (equal number of edges in $T$) valid 2-partitions, six ($=\binom{4}{2}$) valid 3-partitions, and one valid 4-partition.