When specifying the universal property that defines $V \otimes W$, what category are we working in? What category does $V \otimes W$ satisfy its universal property in? I'm having a hard time figuring out what the arrows in the category in question are supposed to be. I don't think it's just the category of vector spaces with arrows being linear maps because bilinear maps are not linear maps and I'm not sure what collection of arrows contains bilinear and linear maps without simply containing all functions between vector spaces.
All vector spaces in this question are over a fixed field $F$.

The universal property of definition of the tensor product is given here on Wikipedia.
I've reproduced a similar diagram to this one below, but added in an extra identity arrow from $Z$ because I can't figure out how to draw a diagonal arrow, also the arrow from $V \otimes W$ to $Z$ should be dashed. I also changed the notation for $h'$ to use a prime instead of a superscript tilde.
In the diagram below, $h$ is bilinear, $\varphi$ is bilinear, and $h'$ is linear. Additionally $h'$ is uniquely determined.
$$
\require{AMScd}
\begin{CD}
V \times W @>{\varphi}>> V \otimes W\\
@V{h}VV @VV{h'}V \\
Z @>{\text{id}}>> Z
\end{CD}
$$
I don't understand which category specifically this diagram is in.
$V$, $W$, $V \times W$, $Z$ and $V \otimes W$ are all vector spaces. Additionally, $h$ and $\varphi$ are bilinear, but $h'$ is linear, so it's not clear to me what the collection of arrows as a whole should be.

What follows is my understanding of how to build $V \otimes W$ by imposing equalities on the free vector space over $V \times W$.
This is my attempt to understand how the tensor product of vector spaces is defined using different machinery.

The concrete construction for the tensor product makes intuitive sense and I think I understand the intuition for why a bilinear map from $V \times W$ to an arbitrary $Z$ would necessarily be completely determined by what it does to the basis of $V \otimes W$. I don't have a proof of this fact at the moment though.
Here's my understanding of how to produce a specific concrete vector space given two vector spaces.
I'll use ${v \choose w}$ to nonstandardly denote the formal tensor product of $v$ and $w$, which I'm thinking of as a fancy kind of ordered pair.
Let $V$ and $W$ be vector spaces. I will now define $T$, the tensor product of $V$ and $W$.
Let $T^1$ be defined as $\{ {v \choose w} \mathop| v \in V \land w \in W \} $.
Let $T^2$ be the set of formal finite linear combinations of elements of $T_1$, with coefficients taken from $F$, our fixed field of scalars.
Let $T^3$ be defined by taking $T^2$ and modding out by the congruence relation $\simeq$.
Let $\simeq$ be the congruence relation generated by $(\simeq_0)$ and respecting the operations $+$ and scalar multiplication by $k$. $(\simeq_0)$ is defined below.
$$ {v_1 \choose w} + {v_2 \choose w} \;(\simeq_0)\; {v_1 + v_2 \choose w} $$
$$ {v \choose w_1} + {v \choose w_2} \;(\simeq_0)\; {v \choose w_1 + w_2} $$
$$ (k){v \choose w} \;(\simeq_0)\; {kv \choose w} $$
$$ (k){v \choose w} \;(\simeq_0)\; {v \choose kw} $$
Let $T$ be equal to $T^3$.
 A: The term "universal property" does not have a universally agreed upon definition, as far as I know. But most universal properties can be understood in terms of representable functors.
Let $C$ be a category. A functor $F\colon C\to \mathsf{Set}$ is representable if there is some object $x\in C$ such that $F$ is naturally isomorphic to the Hom  functor $h^x\colon  C\to \mathsf{Set}$, which is defined on objects by $h^x(y) = \mathrm{Hom}(x,y)$ and on arrows by $h^x(f) = f\circ -\colon \mathrm{Hom}(x,y)\to \mathrm{Hom}(x,z)$ for  $f\colon y\to z$.
If $F$ is representable by $x$, then the set $\mathrm{Hom}(x,y)$ is in natural bijection with the set $F(y)$. That is, arrows out of $x$ to $y$ are classified by elements of $F(y)$.
In fact, the situation is even better than that! The identity arrow $\mathrm{id}_x\in \mathrm{Hom}(x,x)$ corresponds to an element $u\in F(x)$, and $u$ is "universal" in the following sense: For every object $y$ in $C$ and every element $a\in F(y)$, there is a unique arrow $f\colon x\to y$ in $C$ such that $F(f)(u) = a$.
The existence of $u\in F(x)$ with  this universal property is equivalent to the statement that $F$ is representable by $x$ - the proof is essentially the proof of the Yoneda lemma. Of course this whole discussion has  a dual version for universal properties in which maps into $x$ from $y$ are classified  by  elements of $G(y)$ for  a contravariant functor $G$ to $\mathrm{Set}$.

Ok, what about tensor products? Fixing vector spaces $V$ and $W$ over a field $k$, there is a functor $B_{V,W}\colon \mathrm{Vect}_k\to \mathrm{Set}$ defined by $B_{V,W}(U) = \{f\colon V\times W\to U\mid f\text{ is bilinear}\}$. If $T\colon U\to U'$ is a morphism of vector spaces (a linear transformation) and $f\colon V\times W\to U$ is bilinear, then $T\circ f\colon V\times W\to U'$ is bilinear. This defines the action of $B_{V,W}$ on morphisms: $B_{V,W}(T)\colon B_{V,W}(U)\to B_{V,W}(U')$ is defined by $B_{V,W}(T)(f) = T\circ f$.
Now what does it mean to say that this functor $B_{V,W}$ is represented by the tensor product $V\otimes W$? By the discussion above, there is a universal $b\in B_{V,W}(V\otimes W)$ such that for any vector space $U$ and any element $f\in B_{V,W}(U)$, there is a unique morphism of vector spaces $\tilde{f}\colon V\otimes W\to U$ such that $f = B_{V,W}(\tilde{f})(b)$. Unpacking, we have a distinguished bilinear map $b\colon V\times W\to V\otimes W$, and for any bilinear map $f\colon V\times W\to U$, there is a unique morphism of vector spaces $\tilde{f}\colon V\otimes W\to U$ such that $f = \tilde{f}\circ b$. This is exactly the familiar universal property of the tensor product.

Note that there was no need to consider any categories in the above discussion other than the category $\mathrm{Vect}_k$ of vector spaces over $k$ and the category $\mathrm{Set}$ of sets. To define the functor $B_{V,W}$, we did have to consider composing two different kinds of maps, namely bilinear maps and linear maps. But of course these are special kinds of functions, and we know how to compose functions - to make sense of this, we are not required to consider a category where the objects are vector spaces and the arrows include both linear maps and bilinear maps.
Nevertheless, it is reasonable to wonder whether such a category exists. It turns out that the right type of structure to consider here is not a category, but a multicategory. Multicategories are generalizations of categories in which every arrow has a finite sequence of objects as its domain. The main motivating example is the multicategory where the objects are vector spaces and the arrows $(V_1,\dots,V_n)\to W$ are the multilinear maps $V_1\times\dots\times V_n\to W$.
In this multicategory, the tensor product $V_1\otimes\dots\otimes V_n$ has the universal property that any arrow $(V_1,\dots,V_n)\to W$ factors uniquely as the canonical arrow $(V_1,\dots,V_n)\to V_1\otimes \dots\otimes V_n$ followed by an arrow $V_1\otimes \dots\otimes V_n\to W$.
Since I know that you (the OP) are interested in logic, I'll mention that other examples of multicategories come from logic. For example, take objects to be sentences and arrows $(P_1,\dots,P_n)\to Q$ to be sequents $P_1,\dots,P_n\vdash Q$ (so there is an arrow $(P_1,\dots,P_n)\to Q$ if and only if $Q$ follows from assumptions $P_1,\dots,P_n$). In classical propositional logic, the conjunction $P_1\land \dots \land P_n$ has the universal property of the tensor product.
