Your question is a very natural one (if I understand it correctly) and at the same time it raises a rather difficult question.
As you say, it makes sense to identify the operators $A : X \to Y$ of finite rank with elements of $X^{\ast} \otimes Y$. Now you want the operator norm of $A$ to coincide with its norm in $X^{\ast} \otimes Y$ with respect to some tensor norm. I leave it to you to check that the injective tensor norm defined for $\omega = \sum x_{i}^{\ast} \otimes y_{i}$ by
$$\Vert \omega \Vert_{\varepsilon} = \sup_{\substack{\Vert \phi \Vert_{X^{\ast\ast}} \leq 1 \\\ \Vert \psi \Vert_{Y^{\ast}} \leq 1}}{\left\vert \sum \phi(x_{i}^{\ast}) \, \psi(y_{i})\right\vert}$$
does what you want (it is independent of the representation of $\omega$ as a finite sum of elementary tensors and $\|A\| = \|A\|_{\varepsilon}$ for operators of finite rank). Edit: To see the second claim, use Goldstine's theorem that allows you to replace the supremum over $\phi \in X^{\ast\ast}$ with $\Vert\phi\Vert_{X^{\ast\ast}} \leq 1$ by the supremum over $\operatorname{ev}_{x}$ with $x \in X$ and $\Vert x \Vert_{X} \leq 1$.
The projective tensor norm of a finite rank operator is usually much larger than its operator norm (see the discussion on pp.41ff in Ryan, for example).
Given this, we can identify the completion $X^{\ast} \otimes_{\varepsilon} Y$ of $X^{\ast} \otimes Y$ with respect to the injective tensor norm with a space $K_{0}(X,Y) \subset L(X,Y)$ of operators $X \to Y$ and we will freely do so from now on. Note that $K_{0}(X,Y)$ is nothing but the closure of the operators of finite rank in $L(X,Y)$.
Now the question is: What are the operators lying in $K_{0}(X,Y)$ ?
This is really difficult and I'll outline the closest I know to an answer to that.
As a first observation note that the compact operators $K(X,Y) \subset L(X,Y)$ are a closed subspace of $L(X,Y)$ containing $X^{\ast} \otimes_{\varepsilon} Y = K_{0}(X,Y)$. Looking at the examples of Hilbert spaces or the classical Banach spaces one finds out that quite often $K(X,Y) = K_{0}(X,Y)$ holds. However, it may fail in general, and that's where the famous Approximation Property comes in. I'll refrain from delving into the numerous equivalent formulations and use it as a black box. We have the following theorem due to Grothendieck:
Theorem. The Banach space $X^{\ast}$ has the approximation property if and only if $K_{0}(X,Y) = K(X,Y)$ holds for all Banach spaces $Y$.
Edit 2: (in response to a comment of the OP) It follows that for a reflexive Banach space $X$ with the approximation property we have $K(X,Y) = X^{\ast} \otimes_{\varepsilon} Y$ for all Banach spaces $Y$.
Now most of the Banach spaces you'll run into have the approximation property, e.g. $L^{p}$, $C(X)$ and so on. However, P. Enflo (in a veritable tour de force) has shown that there exist Banach spaces failing the approximation property. An explicit example (identified by Szankowski) is the space $L(H,H)$ of a separable Hilbert space. Note that this space is the dual space of the trace class operators. A famously open question is whether the space $H^{\infty}(D)$ of bounded holomorphic functions on the open unit disk has the approximation property.
I hope this answers your question. The approximation property is discussed in detail in any book that treats the tensor products of Banach spaces. In particular, this is well treated in Ryan's book.
Question 1: Well, you can impose that norm, but normally we don't. The whole point of talking about the "algebraic" tensor product is that we are considering just vector spaces, and forgetting about any norms or topology. In fact, the "operator norm" you propose is exactly the same as the weak tensor norm, since what you call $T(f,g)$ is equal to $\sum_{i=1}^{n} f(x_i)g(y_i)$ if $T=\sum x_i\otimes y_j$.
Question 2: Yes, this is obvious, since as remarked above, the expression $\sum_{i=1}^{n} f(x_i)g(y_i)$ is just the evaluation of the bilinear map $u$ on $(f,g)$.
Question 3: It is saying there is only one product with the property that $(x_1 \otimes y_1)(x_2 \otimes y_2) = x_1x_2 \otimes y_1y_2$. Obviously if you don't impose this condition then there are many many different products, since you just have some abstract vector space and no additional constraints.
Question 4: No, $X\otimes Y$ will usually not be a Banach algebra, because it will not be complete with respect to the projective tensor norm. You extend the product to the completion using density: given $a,b\in X\otimes_p Y$, we can write $a$ and $b$ as the limits of sequences $(a_n)$ and $(b_n)$ in $X\otimes Y$, and then define $ab$ to be the limit of $a_nb_n$. (Of course, you must prove that this limit exists and is independent of the choice of sequences, and that this really does make $X\otimes_p Y$ a Banach algebra.) This extension is the unique extension of the product on $X\otimes Y$ to a product on $X\otimes_p Y$ which is continuous with respect to the norm (indeed, the definition above is forced if you want it to be continuous). If you don't care about continuity, the extension will usually not be unique.
Best Answer
It is not necessarily true that $(X \otimes Y)/Y$ is isomorphic to $X$; it may help verify that even in the finite-dimensional case, the dimensions of these spaces do not match.
I think that the easiest approach for the case where $Y$ is finite dimensional is to note that if $Y$ has dimension $n$ (so that $Y$ is isomorphic to $\Bbb C^n$ under some norm), we have $$ X \otimes Y \cong X \otimes \Bbb C^n \cong \overbrace{X \oplus \cdots \oplus X}^{n \text{ times}}. $$ In the infinite-dimensional case, suppose that $\{x_i\} \subset X$ and $\{y_i\} \subset Y$ are linearly independent sequences with $\|x_i\| = \|y_i\| = 1$. It suffices to verify that the sequence defined by $$ z_n = \sum_{i=1}^n \frac{1}{2^i} x_i \otimes y_i $$ is Cauchy, but fails to converge in $X \otimes Y$ (since the infinite sum cannot be expressed as a finite sum of tensor products).