Let $V,W$ be vector spaces over $\mathbb{F}$.
$B=(v_1,...,v_n)$ basis of $V$,$\quad C=(w_1,...,w_m)$ basis of $W$.$U_1=\operatorname{span}(\{v_1,...,v_l\}),\quad U_2=\operatorname{span}(\{v_{l+1},...,v_n\}),\quad 1\leq l\leq n$
$Z_1=\operatorname{span}(\{w_1,...,w_k\}),\quad Z_2=\operatorname{span}(\{w_{k+1},...,w_m\}),\quad 1\leq k\leq m$
$L=\{T\in \operatorname{Hom}(V,W)|\:T(U_1)\subset Z_1\}$
1) Express $\phi(L)\:$ (where $\phi:\operatorname{Hom}(V,W)\to M_{m\times n}(\mathbb{F})$ is defined by $\phi(T)=[T]^B_C$)
2) Find $\dim L$
Progress:
1) I'm trying to understand what it means to express $\phi(L)$ and how should I look at $T(U_1)\subset Z_1$.
2) Since, $\dim L=\dim \phi(L)$ we can find a basis for $\phi(L)$
Here's one way to think about it: note that if $T(u_p) = \sum_{q=1}^m c_{pq} w_q$, then $(c_{p1},\dots,c_{pm})$ is the $p$th column of the matrix $\phi(T)$ (note that these coefficients $c_{pq}$ are uniquely determined because $C$ is a basis).
With that being said: if $T \in L$, then for $p = 1,\dots,\ell$, we have $T(u_p) \in Z_1$, which is to say that we may write $T(u_p) = \sum_{q=1}^k c_{pq} w_q$. That is, the $p$th column of $\phi(T)$ has the form $(c_{p1},\dots,c_{pk},0,\dots,0)$. Because this applies to columns $1$ through $\ell$ of $\phi(T)$, we may partition the matrix $\phi(T)$ as follows: $$ \phi(T) = \left[\begin{array}{c|c}M_{11} & M_{21}\\ \hline 0 & M_{22}\end{array}\right] $$ where $M_{11}$ has size $k \times \ell$. Now, verify that if $\phi(T)$ has the above form, then $T$ is necessarily an element of $L$ (the converse to our conclusion above).
With that, we have characterized $\phi(L)$; as Bernard's comment hints, it consists of the "block upper-triangular matrices" under a fixed partition. The computation of $\dim \phi(L)$ is straightforward.