If so, then this would imply that if $L/K$ is a nontrivial purely inseparable extension, then $L \otimes_{K} L$ cannot be reduced (i.e., it must contain a nonzero nilpotent).
This is true and is straightforward to prove directly, and "purely inseparable" can be replaced with "inseparable." If $L/K$ is inseparable it contains $\alpha \not \in K$ such that $\alpha^p = \beta \in K$. Then $L \otimes_K L$ contains
$$L \otimes_K K[\alpha_R]/(\alpha_R^p - \beta) \cong L[\alpha_R]/(\alpha_R^p - \beta) \cong L[\alpha_R]/(\alpha_R - \alpha_L)^p$$
where $\alpha_R = 1 \otimes \alpha$ and $\alpha_L = \alpha \otimes 1$. It follows that $\alpha_R - \alpha_L$ is a nonzero nilpotent in $L \otimes_K L$.
This is an if-and-only-if; conversely, suppose that $L/K$ is separable. Then $L \otimes_K L$ is the increasing union of $L \otimes_K L'$ where $L'$ runs over all finite subextensions of $L$, and each of these finite separable subextensions has a primitive element so has the form $K[\alpha]/f(\alpha)$ for $f$ a separable polynomial. This gives that $L \otimes_K L' \cong L[\alpha]/f(\alpha)$ which is a product of fields by the Chinese remainder theorem, and in particular reduced. So $L \otimes_K L$ is an increasing union of reduced rings and hence reduced. (Probably there's a better way to do this.) In summary:
Lemma 1: An algebraic extension $L/K$ is separable iff $L \otimes_K L$ is reduced.
It's also true that $L/K$ is separable iff $K^s \otimes_K L$ is reduced where $K^s$ is the separable closure, so equivalently iff $L$ is geometrically reduced, and the proof is basically the same; this may be more intuitive.
Your desired equivalences are also true. It sounds like you already know this but let's state the key fact about epimorphisms clearly. In any category whatsoever, the cokernel pair of a morphism $f : x \to y$ is, if it exists, the pushout $y \sqcup_x y$ of $f$ along itself.
The cokernel pair detects epimorphisms: $f$ is an epimorphism iff $\text{coker}(f)$ exists and is is canonically isomorphic to $y$, in the sense that the pushout $y \sqcup_x y$ exists and the natural map $y \sqcup_x y \to y$ both of whose components are $\text{id}_y$ is an isomorphism, or equivalently iff $\text{id}_y$ exhibits $y$ as the cokernel pair of $f$.
This is just a matter of unwinding definitions; if you work through what this means using the universal property of the pushout it says exactly that $f$ is an epimorphism. Notably this statement requires no hypotheses on the category $C$ and in particular does not require $C$ to have all pushouts.
Corollary 2: A morphism $f : R \to S$ of commutative rings is an epimorphism iff the multiplication map $S \otimes_R S \to S$ is an isomorphism.
This follows from the fact that tensor product computes the pushout of commutative rings, together with the fact that the natural map from the pushout to $S$ is the multiplication map in the case of commutative rings, which is not hard to verify. Now we need to know something about pushouts in the category of fields and of reduced rings (assumed commutative throughout). The case of reduced rings is easier:
Lemma 3: The pushout $B \sqcup_A C$ in the category of reduced rings is the quotient of the tensor product $B \otimes_A C$ by its nilradical. Hence a map $f : A \to B$ of reduced rings is an epimorphism in reduced rings iff the kernel of the multiplication map $B \otimes_A B \to B$ is nilpotent.
This follows from Corollary 2, the fact that the tensor product computes the pushout in commutative rings, and the fact that the inclusion of reduced rings into commutative rings has left adjoint given by quotienting by the nilradical. Now we can prove:
1 is equivalent to 3: A field extension $f : K \to L$ is an epimorphism in reduced rings iff the kernel of multiplication $L \otimes_K L \to L$ is nilpotent iff $L/K$ is purely inseparable.
Proof. By Lemma 3, $f$ is an epimorphism iff the kernel of the multiplication map $L \otimes_K L \to L$ is nilpotent. For every $\alpha \in L$ this kernel contains $\alpha \otimes 1 - 1 \otimes \alpha$, and if $\alpha \not \in K$ then this element is not zero in the tensor product, so must be a nontrivial nilpotent. This means
$$(\alpha \otimes 1 - 1 \otimes \alpha)^n = \sum_{k=0}^n (-1)^{n-k} {n \choose k} \alpha^k \otimes \alpha^{n-k} = 0 \in L \otimes_K L.$$
If $\alpha$ were transcendental then the tensor products $\alpha^k \otimes \alpha^{n-k}$ would all be linearly independent, so $\alpha$ is algebraic over $K$. By Lemma 1, $L/K$ is inseparable, and in particular $K$ has positive characteristic $p$. This gives
$$(\alpha \otimes 1 - 1 \otimes \alpha)^{p^k} = \alpha^{p^k} \otimes 1 - 1 \otimes \alpha^{p^k} = 0$$
for some $k$, so $\alpha^{p^k} \in K$ for some $k$. Since $\alpha$ is arbitrary in this argument, it follows that the kernel of multiplication $L \otimes_K L \to L$ is nilpotent iff $L/K$ is purely inseparable as desired. $\Box$
To finish we need to know something about pushouts in the category of fields.
Lemma 4 (2 is equivalent to 3): The pushout $L_1 \sqcup_K L_2$ in fields exists iff the quotient of $L_1 \otimes_K L_2$ by its nilradical is a field, and is that field. Hence a field extension $f : K \to L$ is an epimorphism in fields iff the kernel of multiplication $L \otimes_K L \to L$ is nilpotent iff it is an epimorphism in reduced rings.
For a proof see here, although it's not hard to deduce from Lemma 3. We can also give a different proof that a morphism of fields is an epimorphism in fields iff it's an epimorphism in reduced rings using the fact that the reduced rings are exactly the rings that embed into a product of fields.
