Let $k$ be a field. For a finite dimensional $k$-vector space $V$, write $k[V]$ for the symmetric algebra, which is noncanonically isomorphic to some polynomial ring in finitely many variables.
Take a short exact sequence of $k$-vector spaces $0 \rightarrow U \rightarrow V \rightarrow W \rightarrow 0$. This induces a surjection $f : k[V] \rightarrow k[W]$. There is a map $U \otimes_k k[V] \rightarrow k[V]$ sending $u \otimes f$ to $u \otimes f$. My question is, what is the kernel of this map, and is the image of this map equal to the kernel of $f$?
Note that $U \otimes k[V] \rightarrow V \otimes k[V]$ is an injection. On each graded piece, $V \otimes S_n (V) \rightarrow S_{n+1} (V)$ is not an injection; we must symmetrize. So the kernel of $U \otimes S_n (V) \rightarrow S_{n+1} (V)$ should consist of relations induced by permuting terms.
Take an element of the kernel of $S_{n+1} (V) \rightarrow S_{n+1}(W)$. For now, let it just be a monomial $v_1 \otimes \cdots \otimes v_n$, so that the monomial $f(v_1) \otimes \cdots \otimes f(v_n)$ is $0$ in $W$. We should be able to say that at least one of the entries in this term is in the kernel of $f$, right?
Let's pick bases so everything is very concrete in terms of polynomials. We may pick a basis for $U$ and extend it to a basis for $V$. Let us write our basis for $U$ as $x_1,\dots,x_n$ and the basis for $V$ as $x_{n+1},\dots,x_{n+m}$ (though no assertion is being made that these bases are finite; this is just convenient notation). We may then consider $x_{n+1},\dots,x_{n+m}$ as a basis for $W$. We have $k[V]=k[x_1,\dots,x_{n+m}]$ and $k[W]=k[x_{n+1},\dots,x_{n+m}]$ and $f:k[V]\to k[W]$ is the map sending $x_1,\dots,x_n$ to $0$. Your map $g:U\otimes k[V]\to k[V]$ is just the multiplication map, when you consider an element of $U$ as a homogeneous linear polynomial in $x_1,\dots,x_{n}$.
In particular, this makes it obvious that the image of $g$ and the kernel of $f$ are both the ideal $(x_1,\dots,x_n)\subseteq k[V]$, so yes, they are equal.
The kernel of $g$ can be described reasonably explicitly. Note that $U\otimes k[V]$ can naturally be considered as a $k[V]$-module and $g$ is $k[V]$-linear, so $\ker g$ is a $k[V]$-submodule of $U\otimes k[V]$. As a $k[V]$-module, $\ker g$ is generated by the elements $x_i\otimes x_j-x_j\otimes x_i$ for all $1\leq i<j\leq n$. Much more broadly, the exact sequence with $f$ and $g$ which you are considering has a natural extension to a free resolution of $k[W]$ as a $k[V]$-module known as the Koszul complex: $$\dots\to \bigwedge^3 U\otimes k[V]\to \bigwedge^2 U\otimes k[V]\stackrel{h}\to \bigwedge^1 U\otimes k[V]\stackrel{g}\to \bigwedge^0 U\otimes k[V]\stackrel{f}\to k[W]\to 0.$$ Here $\bigwedge^0 U\cong k$ so $\bigwedge^0 U\otimes k[V]$ can be naturally identified with $k[V]$, and $\bigwedge^1 U\cong U$ so $\bigwedge^1 U\otimes k[V]$ can naturally be identified with $U\otimes k[V]$. The map labelled $h$ then sends $(x_i\wedge x_j)\otimes 1$ to $x_i\otimes x_j-x_j\otimes x_i$ and so its image (the kernel of $g$) is exactly the submodule of $U\otimes k[V]$ generated by these elements, as described above. More generally, the map $d_r:\bigwedge^{r} U\otimes k[V]\to \bigwedge^{r-1} U \otimes k[V]$ in this sequence is given by $$d_r((x_{i_1}\wedge\dots \wedge x_{i_r})\otimes 1)=\sum_{j=1}^r (-1)^j (x_{i_1}\wedge\dots\wedge \hat{x}_{i_j}\wedge\dots\wedge x_{i_r})\otimes x_{i_j}$$ where the hat indicates $x_{i_j}$ is omitted. You can find more information on this resolution (and how to prove it is exact) at the link above, or in many textbooks on commutative algebra.