You are probably confusing two different ways to derive the equation, one not using the concept of potential, and the other one using the concept.
If we are starting with Faraday's law for a closed path, we do not use the concept of potential, instead we need to express contributions to the integral of total electric field $\mathbf E$
$$
\oint \mathbf E \cdot d\mathbf{l}
$$
for the oriented path colocated with the conductive path of the circuit, due to each conducting member in the circuit.
For the circuit in the picture, we have
$$
\int \mathbf E \cdot d\mathbf l = 0 ~~~\text{(for the inductor)}
$$
We have zero because the inductor is made of perfect conductor, where total electric field vanishes.
$$
\int \mathbf E \cdot d\mathbf l = RI ~~~\text{(for the resistor)}
$$
Here we have plus sign because in resistor, electric field is in direction of current.
$$
\int \mathbf E \cdot d\mathbf l = -\epsilon_c ~~~\text{(for the cell)}
$$
Here we have minus sign, because electric field inside the cell is opposite to direction of current, which is the direction of integration. Current in a cell flows in direction of electrochemical EMF, which points against electric field there.
So the Faraday law equation becomes
$$
RI - \epsilon_c = -L\frac{dI}{dt}.
$$
The other way, used in practice when solving AC circuits, is to write down directly the KVL equation, which states that sum of potential drops in a closed path equals to 0. We have
$$
\text{potential drop on the inductor} = L\frac{dI}{dt},
$$
$$
\text{potential drop on the resistor} = RI,
$$
$$
\text{potential drop on the cell} = -\epsilon_c.
$$
So the KVL equation is
$$
L\frac{dI}{dt} + RI - \epsilon_c = 0.
$$
which is the same as the equation obtained using the first method.