LFSR get output from characteristic polynomial?
The variable of a polynomial is traditionally noted $x$, not $X$; and, when dealing with LFSRs, the polynomial is seldom considered a function. Thus I'll rewrite the polynomial as $P(x)=x^4+x^3+1$, a polynomial of degree $n=4$.

This is a polynomial with coefficients in the field $\operatorname{GF}(2)$, also noted $\mathbb Z_2$ or $(\{0,1\},+,\cdot)$ [note: in $\operatorname{GF}(2)$, $1+1=0$, and $-$ is the same as $+$]. The polynomial's coefficients are all either $0$ or $1$, and not shown: a term $x^j$ is present when its coefficient is $1$, and absent when its coefficient is $0$. When performing polynomial arithmetic, we are manipulating polynomials irrespective of the value and domain of their variable and result, unless otherwise stated.



The state of the LFSR is a polynomial $S(x)$ with coefficients in $\operatorname{GF}(2)$ of degree at most $n-1$, but can also be thought as:


A vector of $n$ bits $(s_0,s_1,\dots,s_{n-1})$ that are the coefficients of $S$ starting from the constant term going up to higher order terms, that is with $S(x)=\sum_{j=0}^{n-1}s_j\cdot x^j$ with the convention that $x^0=1$;
The integer $\hat S=\sum_{j=0}^{n-1}s_j\cdot 2^j$; addition of polynomials becomes bitwise eXclusive-OR, and multiplication of polynomials becomes carry-less multiplication [performed as binary multiplication, but replacing additions with bitwise eXclusive-OR]; the same integer can be defined as $\hat S=S(2)$ [note: here we are considering $S$ as a polynomial function over $\mathbb Z$].


The next state of the LFSR is, by definition, the remainder of the polynomial division of the polynomial $x\cdot S(x)$ by the polynomial $P(x)$, noted $\big(x\cdot S(x)\big)\bmod P(x)$.
By definition of polynomial division, the remainder of the division of polynomial $T(x)$ by non-zero polynomial $P(x)$ is the polynomial $R(x)$ of degree less than $P(x)$ such that there exists a polynomial $Q(x)$ with $T(x)=P(x)\cdot Q(x)+R(x)$.
In the context of moving to the next step of a LFSR, $Q(x)$ is either $Q(x)=1$ [when $S$ is of degree $n-1$] or $Q(x)=0$ [otherwise]; thus the next state $\big(x\cdot S(x)\big)\bmod P(x)$ is either $x\cdot S(x)+P(x)$ [when $S$ is of degree $n-1$] or $x\cdot S(x)$ [otherwise].

For example, if the initial state of the LFSR is $S_0(x)=x^2$,


the first non-initial state is $S_1(x)=\big(x\cdot S_0(x)\big)\bmod P(x)=\big(x\cdot x^2\big)\bmod P(x)=x^3\bmod P(x)=x^3$.
the second non-initial state is $S_2(x)=\big(x\cdot S_1(x)\big)\bmod P(x)$ that is $\big(x\cdot x^3\big)\bmod P(x)=x^4\bmod P(x)=x^3+1$, with the last step because $x^4=P(x)\cdot Q(x)+x^3+1$ with $Q(x)=1$, and we can obtain $x^3+1$ as $x^4+P(x)=x^4+x^4+x^3+1=x^3+1$.
$S_3(x)=\big(x\cdot S_2(x)\big)\bmod P(x)=\Big(x\cdot\big(x^3+1\big)\Big)\bmod P(x)$ that is $\big(x^4+x\big)\bmod P(x)=x^4+x+P(x)=x^3+x+1$.
$S_4(x)=\big(x\cdot S_3(x)\big)\bmod P(x)=\Big(x\cdot\big(x^3+x+1\big)\Big)\bmod P(x)$ that is $\big(x^4+x^2+x\big)\bmod P(x)=x^4+x^2+x+P(x)=x^3+x^2+x+1$.
$S_5(x)=\big(x\cdot S_4(x)\big)\bmod P(x)=\Big(x\cdot\big(x^3+x^2+x+1\big)\Big)\bmod P(x)$ that is $\big(x^4+x^3+x^2+x\big)\bmod P(x)=x^4+x^3+x^2+x+P(x)=x^2+x+1$.
$S_6(x)=\big(x\cdot S_5(x)\big)\bmod P(x)=\Big(x\cdot\big(x^2+x+1\big)\Big)\bmod P(x)$ that is $\big(x^3+x^2+x\big)\bmod P(x)=x^3+x^2+x$.


It is easy to perform the same calculation using either bit vectors, or integers, and that's practice in most computer programs. With integers, we define $\hat P=P(2)$ [that is $\hat P=\text{0x19}$ with our $P(x)$], and the state following $\hat S$ with $0\le \hat S&lt;2^n$ is whichever of $(2\cdot\hat S)\oplus \hat P$ or $2\cdot\hat S$ is less [or, equivalently, less than $2^n$], where $\oplus$ is bitwise eXclusive-OR [operator ^ in C].



One common definition of the output of the LFSR is the coefficient of the term of degree $n-1$ in the states of the LFSR. The constant term is also popular when the reduction polynomial $P(x)$ has a constant term, which is the case in practice. Sometime [especially with the later convention], the initial state is not taken in consideration.

In the above example, the output of the LFSR [including the initial state $S_0(x)=x^2$] is thus
$0$, $1$, $1$, $1$, $1$, $0$, $1$...



Sometime it is used a reflected representation, where state $S(x)$ is represented by the integer $\check S=\sum_{j=0}^{n-1}s_j\cdot 2^{n-1-j}=2^{n-1}\cdot S(1/2)$ [note: here we are considering $S$ as a polynomial function over $\mathbb Q$].
This allows obtaining the output as the low-order bit of the integer representing the state [rather than its bit of rank $n-1$, which might be less easy to isolate]; and makes it possible to compute the next state without any test, using only common operators on unsigned words of size at least $n$ bits, using the C expression
S = (S&gt;&gt;1) ^ (P &amp; -(S&amp;1))
where P is $\check P=\lfloor2^{n-1}\cdot\big(P(1/2)\big)\rfloor$ [that is P of $9$ with our $P(x)$] and fits an unsigned word; &gt;&gt; is unsigned right shift [that's &gt;&gt;&gt; in Java]; &amp; is bitwise AND; and by definition of unary operator - over unsigned words, -U is the unsigned word such that the sum of U and -U is divisible by the smallest power of two that is not representable as an unsigned word [that's the definition of unary operator - using Two's-complement].

For example, starting from the same $S_0(x)=x^2$ and P of $9$, we have S taking the values $2$, $1$, $9$, $13$, $15$, $14$, $7$.. when applying S = (S&gt;&gt;1) ^ (P &amp; -(S&amp;1)) repeatedly, and the same output $0$, $1$, $1$, $1$, $1$, $0$, $1$.. obtained as S&amp;1.



If we are able to do with bit vectors or integers rather than with polynomials, why are we using polynomials, even in theory?

One important reason is that $S_{i+1}=\big(x\cdot S_i(x)\big)\bmod P(x)$ gives $S_i=\big(x^i\cdot S_i(x)\big)\bmod P(x)=\Big(\big(x^i\bmod P(x)\big)\cdot S_i(x)\Big)\bmod P(x)$, and that's key to computing the $i^\text{th}$ state in time $\mathcal O(\log i)$ rather than $\mathcal O(i)$, and linking the period of the generator to properties of $P(x)$.



The schematic in the question (where the first + in reading order is a XOR gate with output on the left, and any other + is just a wire) at first does not seem to be matching what's above in this answer, which applies to the so-called Galois construction of a LFSR. That drawing uses the Fibonacci construction of a LFSR.

The two constructions give the same output, but starting from a different state. The state of a Fibonacci LFSR of degree $n$ [with polynomial $P(x)$ having a constant term] is the first $n$ bits output by the Galois LFSR with the the same polynomial [or that polynomial reflected, depending on definition of a Fibonacci LFSR].

For nice schematics of gate realizations of the two kinds of LFSR, see here.