Given a signature $\Sigma$ and a class of $\Sigma-$algebras $K$, we say that a $\Sigma-$algebra $\textbf{U}=(U, \{\sigma_{\textbf{U}}\}_{\sigma\in\Sigma})$ has the universal mapping property for $K$ over $X\subset U$, which generates $\textbf{U}$, if for every map $f:X\rightarrow A$ admits an extension $\overline{f}:\textbf{U}\rightarrow\textbf{A}$ which is an homomorphism, for all $\textbf{A}=(A, \{\sigma_{\textbf{A}}\}_{\sigma\in\Sigma})$.
What I want to know is if every $\Sigma-$algebra $\textbf{A}$ has the universal mapping property for the class $V(\textbf{A})$ generated by $\textbf{A}$.
I think the answer is yes and here is my reasoning. By Birkhoff's theorem regarding varieties, we know $V(\textbf{A})$ is the class of all $\Sigma-$algebras satisfying the identities $Id(\textbf{A})$ of $\textbf{A}$ over some denumerable set of variables (or at least that is what I remember).
By another theorem by Birkhoff, if $\textbf{T}(X)$ is the algebra of terms over a set $X$ (as a $\Sigma-$algebra, of course), then the quotient $\textbf{F}_{K}(X)=\textbf{T}(X)/ \theta_{K}(X)$ has the universal mapping property for $K$ over $X/\theta_{K}(X)$, for \begin{equation}\theta_{K}(X)=\bigcap\Phi_{K}(X)\end{equation} and \begin{equation}\Phi_{K}(X)=\{\theta\in Con\textbf{T}(X): \textbf{T}(X)/\theta\in IS(K)\},\end{equation} for $Con\textbf{A}$ the set of congruences of the algebra $\textbf{A}$; $I$ the operator associating a class $K$ of $\Sigma-$algebras to the class of all $\Sigma-$algebras that are isomorphic to some element of $K$; and $S$ the same as $I$, but for subalgebras instead of isomorphic ones.
If $K=V(\textbf{A})$, $IS(K)=K$ and $\Phi_{K}(X)$ becomes the set of all congruences of $\textbf{T}(X)$ containing $Id_{X}(\textbf{A})$ (identities on the variables $X$), as far as I understand it, and so $\theta_{K}(X)=Id_{X}(\textbf{A})$.
So $\textbf{T}(X)/Id_{X}(\textbf{A})$ has the universal mapping property for $V(\textbf{A})$ over $X/Id_{X}(\textbf{A})$; but if we take an $X$ such that $|X|\geq|A|$ we can take a trivial surjective map \begin{equation}f:X\rightarrow A\end{equation} that extends to an epimorphism \begin{equation}\overline{f}:\textbf{T}(X)\rightarrow \textbf{A}\end{equation} and by the theorem of isomorphisms $\textbf{T}(X)/Ker\overline{f}\approx\textbf{A}$. Then $\textbf{T}(X)/Ker\overline{f}$ and $\textbf{A}$ have the same identities over $X$, so $Ker\overline{f}=Id_{X}(\textbf{A})$ and therefore $\textbf{A}\approx \textbf{T}(X)/Id_{X}(\textbf{A})$ (this is the step I am most skeptical about, it seems $Ker\overline{f}$ should not equal $Id_{X}(\textbf{A})$ but at the same time I can not point out what its wrong about this reasoning either...).
We can finally conclude that $\textbf{A}$ has the universal mapping property for $V(\textbf{A})$. Is this correct? If so, does a more constructive proof exist? If not, what I did wrong? And if it is incorrect, can anyone provide an example of an universal algebra that never has the universal mapping property?
As mentioned in the comments, the two element lattice is a counterexample to your claim.
As for why your proof is incorrect, your intuition was correct. We can say that $\mathbf{T}(X)/\mathrm{ker}(\bar{f})$ is isomorphic to $\mathbf{A}$ and that they satisfy the same identities, but that only allows us to conclude that $\mathrm{Id}_X(\mathbf{A})\leq \mathrm{ker}(\bar{f})$ (i.e. if $p\approx q$ is an identity of $\mathbf{A}$, then $\bar{f}(p)=\bar{f}(q)$).
More explanation (added in edit): Let $\mathbf{A}$ be the two-element lattice. Let $X=\{x,y\}$. Then $\mathbf{T}(X)/\mathrm{Id}_X(\mathbf{A})$ is the four-element diamond lattice with elements $\{x,y,x\wedge y,x\vee y\}$. If we identify $x$ with the bottom element of $\mathbf{A}$ and $y$ with the top, then $\mathbf{T}(X)/\mathrm{ker}(\bar{f})$ additionally satisfies $x\wedge y\approx x$ and $x\vee y\approx y$. So even though they are both distributive lattices (hence satisfy the same identities) the latter lattice satisfies more relations on the generators (i.e. it is a quotient of the free lattice). So $(x,x\wedge y), (y,x\vee y)\in \mathrm{ker}(\bar{f})$, but $(x,x\wedge y), (y,x\vee y)\not\in\mathrm{Id}_X(\mathbf{A})$.
Further comments: I have typically seen the definition of universal mapping property be the same as yours above except for the assumption that $X$ generates $\mathbf{U}$. When you include that $X$ generates $\mathbf{U}$, that is the definition of an algebra being free over $X$. So your question was actually ''Is every algebra free in the variety it generates?''