Does every universal algebra have the universal mapping property for the class generated by itself? 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? 
 A: 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?'' 
