Jacobson-Morozov theorem for a semisimple algebraic group $G$ (presumably I am working over algebraically closed field) states that: given a unipotent u, there exists a homomorphism $\phi$ from $SL_2$ to $G$ such that the matrix $\begin{pmatrix} 1&1\\0&1\end{pmatrix}$ maps to $u$.
What I want to understand is that when this map is injective and when it has kernel $\pm 1$? Do we know a criteria on $u$ which determines this?
To each unipotent element $u\in G$ one assigns its weighted Dynkin diagram which is basically a map $\Delta\colon\, \Pi\rightarrow \{0,1,2\}$ where $\Pi$ is a basis of simple roots of the root system of $G$. From $\Delta$ one can read off the characters on ${\mathfrak g}={\rm Lie}(G)$ of a maximal torus $T_0$ of an ${\rm SL}_2$ subgroup containing $u$ (all such subgroups are $G$-conjugate in characteristic $0$ by a classical result of Kostant, as David Stewart pointed out in his comment). If $\Delta(\alpha)=1$ for some $\alpha\in \Pi$ then $\mathfrak g$ has a faithful composition factor for the adjoint action of ${\rm SL}_2$. If $\Delta(\Pi)\subseteq \{0,2\}$ then all weights of $T_0$ on $\mathfrak g$ are even which implies that $-1\in{\rm SL}_2$ acts trivially on $\mathfrak g$. So the case of ${\rm PSL}_2$ occurs if and only if $\Delta(\Pi)\subseteq \{0,2\}$. In this case $u$ is sometimes referred to as even. All one has to do now (for $G$ exceptional) is to open Carter's book on groups of Lie type and examine the list of Dynkin labels on pp. 401--406. For $G$ classical it is explained in Carter's book how to construct $\Delta$ from the partition associated with $u$. So it fairly straightforward to figure out which unipotent elements are even for $G$ classical. Although any even unipotent element of $G$ has to be Richardson, there are plenty of them around. For example, all distinguished unipotent elements of $G$ are even.
I'm going to go a little bit out on a limb here and contradict @YCor's comment above; if I make a mistake, presumably someone will point it out.
If I take a $\mathfrak{sl}_2$ representation, you can figure out whether it integrates to one of $SL_2$ or $PSL_2$ just be looking at how $E=\begin{bmatrix}0 & 1\\ 0&0\end{bmatrix}$ acts: in an representation of $PSL_2$, it has only odd length Jordan blocks; in a faithful representation of $SL_2$ there will be at least one Jordan block of even length. Note that exponentiation doesn't change Jordan type, so the same is true for how $u$ acts in the group action.
Thus, it's easy to check which situation you're in by taking any faithful representation of $G$, and computing the Jordan type of $u$ acting on it: if there are all odd parts, then you have $PSL_2$; if you have any even parts, it must be $SL_2$.
The underlying question here is of interest, though I'm not sure how explicitly it has been answered in the literature. However, your header is misleading: the lemma/theorem of Jacobson-Morozov just concerns the embedding of (nonzero) nilpotent elements of a semisimple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ into copies of the unique simple 3-dimensional Lie algebra $\mathfrak{sl}_2$. (The history of the independent work done by Jacobson and Morozov is rather complicated, but anyway there is a modern treatment in N. Bourbaki, Groupes et algebres de Lie, Chap. VIII, $\S11$, among other places. Work of Kostant and others refined the ideas further.)
It then follows from Jacobson-Morozov along with some basic Lie theory that a corresponding Lie group homomorphism $\phi:\mathrm{SL}_2 \rightarrow G$ exists, as in the question. Here $u$ is a nontrivial unipotent element obtained from the given nilpotent element in $\mathfrak{g} = \mathrm{Lie}(G)$ by something like exponentiation. Your problem is to distinguish between embeddings of $\mathrm{SL}_2$ and $\mathrm{PSL}_2$ (equal to $\mathrm{PGL}_2$ over an algebraically closed field).
Actually, one only needs an algebraically closed field of characteristic 0 here, since semisimple Lie groups are the same as semisimple algebraic groups, while there has been careful study of prime characteristic analogues by many people to carry over some of the ideas. Moreover, the study of simple algebraic groups or Lie algebras is usually sufficient in characteristic 0.
The group $G$ has a faithful (rational) representation, though for $G$ of type $E_8$ (a useful test case since the abstract group is simple) the smallest dimension possible is 248 (adjoint representation). In all characteristics the Jordan block decomposition of each unipotent element in such a representation for each exceptional Lie type has been heroically worked out by R. Lawther in Comm. Algebra 23 (1995), 4125-4156. Since there 69 nontrivial unipotent classes for type $E_8$, the resulting tables are nontrivial to compute. Typically there are some Jordan blocks of odd size (or 1), but these seem to be compatible with representations of $\mathrm{SL}_2$, which has irreducible representations of all possible dimensions $\geq 1$. It's true that if some Jordan block is of even size, this is the only possible type of subgroup.
