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Let $T$ be a simple theory. We work in a monster model $\mathbb{M}$ of $T$. Now the following result is well known:

Let $p(x)$ be a type over $A$. Then $p$ does not fork over $A$.

Now suppose that $a\models{p}$. Now the above seems to imply that $a \overset{\vert}{\smile}_{A}A$. the usual statement for the existence of non-forking ( the proof which depends on this lemma) would yield that there is some $a'$ with $\text{tp}(a'/A)=\text{tp}(a/A)$ such that $a'\overset{\vert}{\smile}_{A}A$. But I am unable to come up with a counterexample where ${a\overset{\not\vert}{\smile}_{A}A}$ but $a' \overset{\vert}{\smile}_{A}A$. Am I interpreting the theorem correctly or am I missing a counter example?

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It's true that for all $a$ and all $A$, $a\underset{A}{\overset{\vert}{\smile}} A$.

The theorem about existence of nonforking extensions says that for all $a$, $A$, and $B$, there is some $a'$ with $\mathrm{tp}(a/A) = \mathrm{tp}(a'/A)$ such that $a'\underset{A}{\overset{\vert}{\smile}} B$.

Of course in the trivial case that we already have $a\underset{A}{\overset{\vert}{\smile}} B$, then we can just take $a' = a$. This is what happens when $A = B$.

Did I understand your question correctly? If not, maybe you can re-explain what your confusion is, and I'll edit this answer.

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    No. It is as I thought it should be. The proof of existence does show that the automorphism comes in because given $A\subseteq{B}$, the way we extend a nf type over $A$ to nf type over $B$ leaves open the possibility that we may need an $a'$ . – 2017-01-22