Are the so-called representations of the Lorentz group actually all representations of it? Fermionic fields change sign under a rotation by $2\pi$. However, in $SO^+\left(1,3\right)$ a rotation by $2\pi$ is the identity. For any representation $R$ of $SO\left(1,3\right)$ then we have 
$$R\left(\Lambda_1\right)R\left(\Lambda_2\right)=R\left(\Lambda_1\Lambda_2\right)$$
taking $\Lambda_1=\Lambda_2=$rotation by $2\pi=$ the identity, we have 
$$R\left(1\right)R\left(1\right)=R\left(1\right)$$
giving $R\left(1\right)=1\neq-1$. Hence, these fermionic fields do not transform under a representation of $SO^+\left(1,3\right)$. Of course, fermions do transform under  a representation of the double cover of $SO^+\left(1,3\right)$, $SL\left(2,\mathbb{C}\right)$, but by the above argument these are not representations of the Lorentz group. 
I feel like my professors talk as if this is not the case, calling them all representations of the Lorentz group. I just wanted to check that this is correct, and they are not really representations of the Lorentz group?
 A: Before answering the question, I'll review the definition of "representation" so that the answer will be clear. If $G$ and $H$ are groups, then a homomorphism from $G$ to $H$ is a map that assigns each element $g\in G$ to some element $\sigma(g)\in H$ such that the basic group operations are respected, such as $\sigma(g_1)\sigma(g_2)=\sigma(g_1 g_2)$ and $\sigma(1)=1$ where $1$ is the identity element in $G$ or $H$, respectively.
A matrix representation of a group $G$ is a homomorphism $\sigma:G\to H$ where $H$ is a group of matrices (usually over $\mathbb{C}$) using matrix multiplication as the multiplication rule for the group. This answer will focus on matrix representations, as implied by the OP.
By this definition, the spin-$1/2$ "representation" of the rotation group $SO(3)$ is not a representation, for precisely the reason noted in the OP. It is a legitimate reprentation of the covering group $SU(2)$, again as noted in the OP. Here I'm using just the rotation group $SO(3)$ as an example instead of the Lorentz group $SO(1,3)$, but the idea is the same.
If $F$ is a covering group of $G$, then a representation of $F$ can be regarded as a projective representation of $G$. Intuitively, if $F$ reduces to $G$ when certain differences between elements of $F$ are ignored (technically, when $G$ is regarded as a quotient of $F$), then a representation of $F$ gives a "representation" (in scare quotes) of $G$ if certain differences bewteen representation-matrices are ignored. The word "projective" refers to ignoring these factors (like factors of $-1$ in the OP's example).
On the other hand, if $G$ is a Lie group, then $G$ has an associated Lie algebra, and we can also talk about representations of the Lie algebra. Since $SO(3)$ and $SU(2)$ have the same Lie algebra (just like $SO(1,3)$ and $SL(2,\mathbb{C})$ have the same Lie algebra), a representation of the Lie algebra of $SU(2)$ is also a legitimate representation of the Lie algebra of $SO(3)$.
So the OP's suspicion is correct: people often use the language carelessly (or at least in non-standard ways). People might say "representation of the group $G$" when they really mean "projective reprentation of the group $G$" or "representation of a covering group of $G$" or "representation of the Lie algebra corresponding to $G$."
