If we take a vector $A$, which has three components, my understanding is that we can write this using Einstein notation as $A_{u}$ where this is actually $A_1+A_2+A_3$.
No. $A_\mu$ is the $\mu^{th}$ component of the covector $A$. Summation is only implied if an index is repeated, e.g. $A_\mu B^\mu=A_0B^0 + A_1B^1+\ldots$
We can also write $g^{\mu\nu}A_\nu = A^\mu$.
Yes. To each covector $A$ with components $A_\nu$ we can associate a vector "partner" $A^\sharp$ with components $(A^\sharp)^\mu := g^{\mu\nu} A_\nu$. It is conventional to drop the $\sharp$ and simply write $A^\mu$, where we distinguish between the components of $A$ and $A^\sharp$ by the placement of the index. Similarly, to each vector $V$ with components $V^\mu$ we can associate a covector $V^\flat$ with components $(V^\flat)_\mu := g_{\mu\nu} V^\nu$.
This can be confusing, because $A$ and $A^\sharp$ are emphatically not the same thing (they aren't even the same type of thing - one is a covector and one is a vector). They are partners, whose partnership is in this case defined by the metric tensor (though any non-degenerate bilinear form would do).
Based on some previous questions you've asked, I believe the convention used by your source lumps vectors and covectors together, treating them as elements of a single vector space. In this convention, $A^\mu$ and $A_\mu$ are called the contravariant and covariant components of the vector $A$, corresponding to the expansion of $A$ in the basis $\{\hat e_\mu\}$ or the dual basis $\{\hat \epsilon^\mu\}$, which is defined such that $\hat\epsilon^\mu \cdot \hat e_\nu = \delta^\mu_\nu$. In other words,
$$A = A^\mu \hat e_\mu = A_\nu \hat \epsilon^\nu$$
The components $A^\mu$ and $A_\mu$ are different because the bases $\{\hat e_\mu\}$ and $\{\hat \epsilon^\mu\}$ are generally different (though they coincide if $\{\hat e_\mu\}$ is an orthonormal basis).