The answer is no, because in general there can be more proper coverings than partitions, which will prevent any injective mapping from coverings to partitions.

For example, if $X$ is countably infinite, then there are only
continuum many partitions, but I claim that there are
$2^{\frak{c}}$ many proper coverings. So there can be no such
injective map from coverings to partitions.

There are continuum many partitions of a countably infinite set
$X$, since every partition is determined by a function
$X\to\omega$.

To see that there are $2^{2^{\aleph_0}}$ many distinct proper
coverings, let me identify $X$ with the nodes of the binary
branching tree $2^{<\omega}$. Consider any collection of branches
through the tree, which covers the tree. Any such collection is a
proper covering, since no branch is contained in another, unless
they are equal. How many proper coverings have this special form?
Well, we can cover the tree with countably many branches, and then
add any further set of branches. Since there are continuum many
branches, this gives rise to $2^{\mathfrak{c}}$ many proper
coverings.

Finally, although you asked for a lattice homomorphism from
$\text{Cov}(X)$ to $\text{Part(X)}$, I claim that the collection of
proper coverings is not generally a lattice. To see this, consider
again the case where $X$ consists of the finite binary sequences,
which can be arranged into the tree $2^{<\omega}$. Let $U_0$ be the
covering consisting of all the eventually-0 branches, and let $U_1$
consist of the eventually-1 branches. These are both proper
coverings of $X$. But I claim there is no greatest lower bound of
$U_0$ and $U_1$. Suppose that $W$ is a proper covering that refines
both $U_0$ and $U_1$. So every element of $W$ is contained in an
eventually-0 branch and also contained within an eventually-1
branch. Thus, every element of $W$ must be finite and linearly
ordered as in the tree. Now, there can be no largest such $W$ below
$U_0$ and $U_1$, because I can glue together any two such finite
sets, as long as they remain linearly ordered, and get a strictly
larger covering (with respect to refinement) which remains below
both $U_0$ and $U_1$. So it is not a lattice. The non-lattice counter-example can be generalized to any infinite set, simply by working on a countably infinite subset.