Mathematical Name for Physical Gauge Symmetries In physics, when talking about a gauge transformation, we always mean two combined transformations. For example, a $U(1)$ gauge transformation is a combination of 

$$ \psi \rightarrow e^{ia(x)} \psi $$

plus a simultaneous transformation on a different function

$$ A_\mu \rightarrow  A_\mu + i  \partial_\mu a(x) $$

Is there some mathematical term for this kind of combined transformation. Or is is something of the form

$$ U(1) \oplus R  \quad  ?$$
Let me first write this a bit differently, you are probably familiar with this.
If we consider gauge transformations $U(x)$ with an arbitrary gauge group $G$, the fields transforms as follows:
$$ \psi \mapsto U \psi \;, \quad A_\mu \mapsto U A_\mu U^{-1} + i (\partial_\mu U) U^{-1} \;. $$
(For $G = U(1)$ if you insert $U = e^{ia}$, you get your formulae back.)

Global Transformations:
Let us first pretend that the second term in the transformation of $A$ was not there, i.e. $A_\mu \mapsto U A_\mu U^{-1}$.
This is what happens for global gauge transformations where $U$ is constant.

Our notation hides the fact that $\psi$ actually transforms in some representation $R$ of $G$:
$\psi$ lives in some vector space $V$ and $U$ is an element of $G$, so when we write $U\psi$ we implicitly mean that $U$ acts on $\psi$ in that representation $R$.
(We do this because usually $G = SU(n)$ and $V = \mathbb C^n$ where there is a natural representation.)

The fields $A_\mu$ take values in the Lie algebra $\mathfrak g$ of $G$ and $G$ acts in the adjoint representation $Ad$ defined by $Ad(g)(A) = g A g^{-1}$.
What I want to show with all of this is:
You were on the right track.
Because the two fields $\psi$ and $A$ transform simultaneously, they transform in the direct sum of the two representations:
$$ R \oplus Ad $$

Gauge Transformations:
Let's consider now the complete picture:
The transformation behavior of $A$ is quite peculiar because the action $A_\mu \mapsto U A_\mu U^{-1} + i (\partial_\mu U) U^{-1}$ is not linear and therefore not a representation at all.
It is better to talk about this in completely different terms (literature recommendation: this link, and also chapters 9 and 10 of Nakahara).

A gauge transformation is actually a diffeomorphism of the principal $G$-bundle $P$.
The fermion field is a section of the vector bundle associated with $P$ over the representation $R$.
A gauge transformation acts naturally on the associated vector bundle, this can be described by a function $U$ like above in a local trivialization.

A connection on the principal bundle induces a connection on the associated vector bundle.
Locally, this connection is described by the vector potential $A$, which is a $\mathfrak g$-valued 1-form.
The connection on $P$ has a very natural transformation behavior under gauge transformation -- the transformation of the vector potential of $A$ results from this.
Let me first write this a bit differently, you are probably familiar with this.
If we consider gauge transformations $U(x)$ with an arbitrary gauge group $G$, the fields transforms as follows:
$$ \psi \mapsto U \psi \;, \quad A_\mu \mapsto U A_\mu U^{-1} + i (\partial_\mu U) U^{-1} \;. $$
(For $G = U(1)$ if you insert $U = e^{ia}$, you get your formulae back.)

Global Transformations:
Let us first pretend that the second term in the transformation of $A$ was not there, i.e. $A_\mu \mapsto U A_\mu U^{-1}$.
This is what happens for global gauge transformations where $U$ is constant.

Our notation hides the fact that $\psi$ actually transforms in some representation $R$ of $G$:
$\psi$ lives in some vector space $V$ and $U$ is an element of $G$, so when we write $U\psi$ we implicitly mean that $U$ acts on $\psi$ in that representation $R$.
(We do this because usually $G = SU(n)$ and $V = \mathbb C^n$ where there is a natural representation.)

The fields $A_\mu$ take values in the Lie algebra $\mathfrak g$ of $G$ and $G$ acts in the adjoint representation $Ad$ defined by $Ad(g)(A) = g A g^{-1}$.
What I want to show with all of this is:
You were on the right track.
Because the two fields $\psi$ and $A$ transform simultaneously, they transform in the direct sum of the two representations:
$$ R \oplus Ad $$

Gauge Transformations:
Let's consider now the complete picture:
The transformation behavior of $A$ is quite peculiar because the action $A_\mu \mapsto U A_\mu U^{-1} + i (\partial_\mu U) U^{-1}$ is not linear and therefore not a representation at all.
It is better to talk about this in completely different terms (literature recommendation: this link, and also chapters 9 and 10 of Nakahara).

A gauge transformation is actually a diffeomorphism of the principal $G$-bundle $P$.
The fermion field is a section of the vector bundle associated with $P$ over the representation $R$.
A gauge transformation acts naturally on the associated vector bundle, this can be described by a function $U$ like above in a local trivialization.

A connection on the principal bundle induces a connection on the associated vector bundle.
Locally, this connection is described by the vector potential $A$, which is a $\mathfrak g$-valued 1-form.
The connection on $P$ has a very natural transformation behavior under gauge transformation -- the transformation of the vector potential of $A$ results from this.