Yes, that's true. Let $f_i : V_i \to W_i$ be two linear maps. Since $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $W_1 \otimes W_2$, we may assume that $f_1,f_2$ are surjective. But then they are split, so that we can assume that $V_i = W_i \oplus U_i$ and that $f_i$ equals the projection $V_i \to W_i$, with kernel $U_i$. Then $V_1 \otimes V_2 = W_1 \otimes W_2 \oplus W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2$ and $f_1 \otimes f_2$ equals the projection of $V_1 \otimes V_2$ onto $W_1 \otimes W_2$. Hence the kernel is $W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2 = U_1 \otimes V_2 + V_1 \otimes U_2$.
This shows even more: The kernel is the pushout $(\ker(f_1) \otimes V_2) \cup_{\ker(f_1) \otimes \ker(f_2)} (V_1 \otimes \ker(f_2))$.
By the way, this argument is purely formal and works in every semisimple abelian $\otimes$-category. What happens when we drop semisimplicity, for example when we consider modules over some commutative ring $R$? Then we only need some flatness assumptions:
Let $f_1 : V_1 \to W_1$ and $f_2 : V_2 \to W_2$ be two morphisms in an abelian $\otimes$-category (for example $R$-linear maps between $R$-modules). If $f_1,f_2$ are epimorphisms, then we have exact sequences $\ker(f_1) \to V_1 \to W_1 \to 0$ and $\ker(f_2) \to V_2 \to W_2 \to 0$. Applying the right exactness of the tensor product twice(!), we get that then also
$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2 \to 0$
is exact. If $f_1,f_2$ are not epi, we can still apply the above to their images and get the exactness of
$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to \mathrm{im}(f_1) \otimes \mathrm{im}(f_2) \to 0.$
Now assume that $\mathrm{im}(f_1)$ and $W_2$ are flat. Then $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $\mathrm{im}(f_1) \otimes W_2$ which embeds into $W_1 \otimes W_2$. Hence, we have still that the sequence
$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2$
is exact. In other words, we have a sum decomposition
$$\ker(f_1 \otimes f_2) = \alpha(\ker(f_1) \otimes V_2) + \beta(V_1 \otimes \ker(f_2)),$$ where $\alpha : \ker(f_1) \otimes V_2 \to V_1 \otimes V_2$ and $\beta : V_1 \otimes \ker(f_2) \to V_1 \otimes V_2$ are the canonical morphisms. In general, these are not monic! However, this is the case, by definition, when $V_1$ and $V_2$ are flat. So in this case we can safely treat $\alpha$ and $\beta$ as inclusions and write $$\ker(f_1 \otimes f_2) = V_1 \otimes \ker(f_2) + \ker(f_1) \otimes V_2.$$
If $f_1,f_2,\dots,f_m$ are linearly independent, then you can complete them to a basis $f_1,\dots,f_m,f_{m+1},\dots,f_n$ of $V^*$. Such a basis is the dual of a basis $e_1,\dots,e_n$ of $V$, see https://math.stackexchange.com/a/1772676/62967.
Now $F(e_1)=(1,0,\dots,0)$ and similarly for $F(e_i)$, $1\le i\le m$. Thus $F$ is surjective, because its range contains a basis.
Let's try the converse. If $f_1$ is a linear combination of $f_2,\dots,f_m$, then
$$
\bigcap_{i=1}^m\ker f_i=\bigcap_{i=2}^m \ker f_i
$$
One inclusion is obvious, namely $\subseteq$, but you proved the other. Now you can consider the linear map
$$
G\colon V\to \mathbb{k}^{m-1},\qquad G(x)=(f_2(x),\dots,f_m(x))
$$
and the rank nullity theorem tells you that
$$
\dim\ker G=n-\dim\operatorname{Im}G\ge n-(m-1)=(n-m)+1>n-m
$$
a contradiction, because
$$
\ker G=\bigcap_{i=2}^m\ker f_i=\bigcap_{i=1}^m \ker f_i
$$
has dimension $n-m$ by assumption.
Best Answer
Take $V=K$ and $W=K^2$; consider $f\colon V\to W$ given by $f(1)=(1,0)$ and $f_1\colon V\to W$ given by $f_1(1)=(0,1)$. Of course $\ker f_1\subseteq\ker f$, but $f$ is not linearly dependent on $f_1$.
Let's look at the general case. If $U=\bigcap_{i=1}^n\ker f_i$, then your maps induce linear maps $\bar{f}_i\colon V/U\to W$ $(i=1,\dots,n)$ and $\bar{f}\colon V/U\to W$. The problem is now whether any linear map $g\colon V/U\to W$ can be obtained as a linear combination of the given ones.
Thus we can assume $U=0$ and that the maps are linearly independent, so the problem becomes
What we can say is that there is an embedding
$$ V=\frac{V}{\bigcap_{i=1}^n\ker f_i}\to\bigoplus_{i=1}^n \frac{V}{\ker f_i}. $$
so $\dim V\le n\dim W$. Therefore, increasing the dimension of $V$ in the original problem doesn't help in the "normalized" situation. This shows also that the assertion is, in general, false: the span of the given maps has dimension $n$, while $\dim\mathrm{Hom}(V,W)=(\dim V)(\dim W)$. You should be able to show a counterexample, now.
Notice that, in the case when $W=K$, it suffices to show that the span of the given map has dimension $\dim V$, which follows from the above embedding.