Expressing linear transforms using linear functionals: is this possible? Work over a fixed but arbitrary field.
Let $Y$ and $X$ denote finite-dimensional vectorspaces, and let $y \in Y^n$ denote a sequence of elements of $Y$, where $n$ is a natural number.
It seems likely that for all linear transforms $\varphi : Y \leftarrow X$ such that the image of $\varphi$ is included in the span of $y$, there exists a sequence $f \in (X^*)^n$ such that $$\mathop{\forall}_{x \in X}\left\{\varphi(x) = \sum_{i\in \mathbb{N}_{<n}}f_i(x)y_i\right\}$$
It also seems likely that if $y$ is linearly independent, then the sequence $f$ is unique.

Question. Is any of this true? If so: how do we know that? (proof request)

 A: There's a general result which helps to see this quite easily. For a compact operator $T:H_1 \rightarrow H_2 $ from one Hilbert space into another, there's the spectral representation theorem: 
$$Tx = \sum_{k = 1}^\infty \lambda_k \langle x ,e_k\rangle f_k,$$
where $\lambda_k$ is the sequence of eigenvalues of $|T|= (T^*T)^{\frac{1}{2}}$ and $e_k,f_k$ are series of vectors in $H_1$ and $H_2$, respectively.
For the finite-dimensional case note that we can identify each of your vector spaces with the Hilbert space $\mathbb C^n$ of appropriate dimension with the standard scalar product. Furthermore, the series above of eigenvalues is in fact just sum a with $n$ summands, as $|T|$ has $n$ eigenvalues. The functionals you're looking for are the maps $x \rightarrow \langle\cdot,e_k\rangle$.
There are surely also other ways, such as representing the linear map as a matrix and writing the matrix as a product of appropriate vectors.
A: Clearly we can assume that $y$ spans $Y$ (otherwise replace $Y$ with $Z = \operatorname{span} \{ y_i : 0 \leqslant i < n\}$). Then some subsequence of $y$ is a basis of $Y$. Without loss of generality, suppose that $\{ y_i : 0 \leqslant i < m\}$ is a basis of $Y$. Let $\{\psi_i : 0 \leqslant i < m\}$ be the dual basis of $\{ y_i : 0 \leqslant i < m\}$, so we have $\psi_i(y_j) = \delta_{ij}$ with the Kronecker-$\delta$. Then define
$$f_i = \psi_i \circ \varphi$$
for $0 \leqslant i < m$ and $f_i = 0$ for $m \leqslant i < n$. Since $\{ y_i : 0 \leqslant i < m\}$ is a basis of $Y$, we can write
$$\varphi(x) = \sum_{j = 0}^{m-1} c_j(x)\cdot y_j\tag{1}$$
in a unique way for every $x\in X$, and hence we have
$$f_i(x) = \psi_i(\varphi(x)) = \psi_i\Biggl(\sum_{j = 0}^{m-1} c_j(x)y_j\Biggr) = \sum_{j = 0}^{m-1} c_j(x)\psi_i(y_j) = \sum_{j = 0}^{m-1} c_j(x)\delta_{ij} = c_i(x),$$
so indeed
$$\varphi(x) = \sum_{i = 0}^{m-1} f_i(x)y_i = \sum_{i = 0}^{n-1} f_i(x)y_i,$$
as desired. The uniqueness of the coefficients in $(1)$ shows that the representation
$$\varphi = \sum_{i = 0}^{n-1} f_i\otimes y_i$$
is indeed unique if $y$ is linearly independent (that is, is a basis of its span). Of course the representation is not unique if $y$ is linearly dependent.
Note that only the finite-dimensionality of $Y$ (resp. $Z$) was used, whether $X$ is finite-dimensional or not plays no role.
